The-ratio of their gravitational potential energy is [tex]1:2[/tex]. The unit of measurement for any and all types of energy, which would include kinetic and potential energy, is kilograms per square second, or kg*m2/s2 (J).
Height [tex]H_{1} = h[/tex]
Height [tex]F_{2} = 2h[/tex]
Mass of body 1 = m
Mass of body 2 = m
Gravitational potential energy of body [tex]1 = mgH 1 =mgh[/tex]
Gravitational potential energy of Body 2 [tex]= mgH 2 =mg(2h)[/tex].
Ratio of gravitational potential energies
[tex]=\frac{mgh}{mg (2h)} =\frac{mgh}{2mgh} =\frac{1}{2} = 1:2[/tex]
What Unit of Measurement Is Potential Energy in?
Due to the fact that energy and work quantify identical types of force, the Joule is indeed the proven scientific measurement device for either of those.
How do you calculate kinetic energy?
K E = 1 2 m v 2 is the formula for kinetic energy. When m denotes the body's weight as well as v denotes its velocity, KE stands for kinetic energy.
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How long would it take a roller skater to travel 25 km if his/her speed is 8 km/hr?
At a pace of 8 km/hr, it would take the roller skater 3.125 hours to complete 25 km.
To determine how long it would take a roller skater to travel 25 km if their speed is 8 km/hr, we can use the formula:
time = distance / speed
In this case, the distance is 25 km and the speed is 8 km/hr. Substituting these values into the formula, we get:
time = 25 km / 8 km/hr
Simplifying, we get:
time = 3.125 hours
Therefore, it would take the roller skater 3.125 hours to travel 25 km at a speed of 8 km/hr. It's important to note that this is assuming a constant speed throughout the journey, and not taking into account any breaks or rest stops the skater may need to take.
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calculate the minimum power density measurable if the detectable photogenerated current is 10% of the dark current.
The minimum power density measurable is 0.1 times the dark current density divided by the elementary charge and the external quantum efficiency of the detector.
The minimum power density measurable can be calculated using the following formula:
P = Jsc / (q * η)
where P is the power density, Jsc is the short-circuit current density, q is the elementary charge, and η is the external quantum efficiency of the detector.
Assuming that the photogenerated current is 10% of the dark current, the total current density (J) can be expressed as:
J = Jdark + Jph
where Jdark is the dark current density and Jph is the photogenerated current density.
Since Jph is 10% of Jdark, we can express Jph as:
Jph = 0.1 * Jdark
The short-circuit current density (Jsc) can be approximated as Jph because the detector is operated under short-circuit conditions.
Thus, Jsc ≈ Jph = 0.1 * Jdark
Substituting this into the formula for P, we get:
P = Jsc / (q * η) ≈ Jph / (q * η) = 0.1 * Jdark / (q * η)
Therefore, the minimum power density measurable is 0.1 times the dark current density divided by the elementary charge and the external quantum efficiency of the detector.
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An alpha particle is a nucleus of doubly ionized helium . It has mass of 6.68×10^- 27 kg and charge of 3.2 ×10^- 19 C. Compare the force of electrostatic repulsion between the two Alpha particles with the force of gravitational attraction between them.
[ where the Universe of gravitational constant = 6.67×10^-11 Nm²/kg² and permittivity of free space=8.85×10^-12 F/m ]
show with steps:'(
Answer: To compare the force of electrostatic repulsion and gravitational attraction between two alpha particles, we need to calculate both forces using the given information:
Electrostatic force of repulsion between two alpha particles:
Coulomb's law states that the force of electrostatic repulsion between two charged particles is given by:
F = (kq1q2)/r^2
where k is Coulomb's constant, q1 and q2 are the charges of the two particles, and r is the distance between them.
Here, q1 = q2 = 3.2×10^-19 C (charge of alpha particle), and r = distance between two alpha particles.
We know that two alpha particles are doubly ionized helium nuclei, so they each contain two protons. Thus, their separation distance will be approximately equal to the sum of their radii, which is about 4 fm (4 × 10^-15 m).
Plugging in these values, we get:
F_elec = (8.85×10^-12 * (3.2×10^-19)^2) / (4×10^-15)^2
F_elec = 8.17×10^-28 N
Gravitational force of attraction between two alpha particles:
The force of gravitational attraction between two particles is given by:
F = G*(m1*m2)/r^2
where G is the gravitational constant, m1 and m2 are the masses of the two particles, and r is the distance between them.
Here, m1 = m2 = 6.68×10^-27 kg (mass of alpha particle), and r = distance between two alpha particles.
Plugging in these values, we get:
F_grav = (6.67×10^-11 * (6.68×10^-27)^2) / (4×10^-15)^2
F_grav = 3.71×10^-57 N
Therefore, we can see that the force of electrostatic repulsion between two alpha particles is much greater than the force of gravitational attraction between them.
F_elec/F_grav = (8.17×10^-28 N) / (3.71×10^-57 N) ≈ 2.2×10^29
Explanation:
Answer: The electrostatic force of repulsion between the two alpha particles is about 5.5 × 10^48 times stronger than the gravitational force of attraction between them.
Explanation:
The force of electrostatic repulsion between two alpha particles can be calculated using Coulomb's law:
F_e = k * (q1 * q2) / r^2
where k is the Coulomb constant (1/4πε0), q1 and q2 are the charges of the particles (in coulombs), and r is the distance between them (in meters).
For alpha particles, the charge is 3.2 × 10^-19 C and the distance between them can be assumed to be the sum of their radii, which is approximately 2 x 10^-15 m. Using the given value of the Coulomb constant and the values for charge and distance, we can calculate the force of electrostatic repulsion:
F_e = (9 × 10^9 Nm²/C²) * [(3.2 × 10^-19 C)^2 / (2 × 10^-15 m)^2]
= 1.15 × 10^-28 N
The force of gravitational attraction between two alpha particles can be calculated using Newton's law of gravitation:
F_g = G * (m1 * m2) / r^2
where G is the gravitational constant (6.67×10^-11 Nm²/kg²), m1 and m2 are the masses of the particles (in kg), and r is the distance between them (in meters).
For alpha particles, the mass is 6.68 × 10^-27 kg, and the distance between them can be assumed to be the sum of their radii, which is approximately 2 x 10^-15 m. Using the given value of the gravitational constant and the values for mass and distance, we can calculate the force of gravitational attraction:
F_g = (6.67 × 10^-11 Nm²/kg²) * [(6.68 × 10^-27 kg)^2 / (2 × 10^-15 m)^2]
= 2.1 × 10^-77 N
Comparing the two forces, we see that the electrostatic force of repulsion is much stronger than the gravitational force of attraction:
F_e / F_g = (1.15 × 10^-28 N) / (2.1 × 10^-77 N) = 5.5 × 10^48
Therefore, the electrostatic force of repulsion between the two alpha particles is about 5.5 × 10^48 times stronger than the gravitational force of attraction between them.
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A point charge (q1) has a magnitude of 3x10-6 C. A second charge (q2) has a magnitude of -1. 5x10-6 C and is located 0. 12 m from the first charge. Determine the electrostatic force each charge exerts on the other
The electrostatic force each charge exerts on the other is 2.81N .[tex]q_{1} =[/tex] [tex]3[/tex] × [tex]10^{-6} C[/tex][tex]\\q_2}[/tex] [tex]= - 1.5[/tex] × [tex]10^{-6} C[/tex] They are separated by a distance [tex]r = 0.12m[/tex]
Electrostatic force is [tex]F =[/tex] [tex]\frac{kq_{1} q_{2} }{r^{2} }[/tex]
[tex]F = 2.81N[/tex]
The term "scalar" refers to a category of quantities that may be fully defined by a single magnitude. Quantities called vectors can be fully represented by their direction as well as their magnitude.
How do you define a vector's magnitude?
A vector's length is considered to be its magnitude. The letter "a" stands for the dynamic array magnitude. For further information about a vector's magnitude, go to its introduction. This article deduces formulas for vectors magnitude in considerations of their coordinates in three and two dimensions, respectively.
Magnitude is simply "distance or amount," according to the definition given in physics. It illustrates the either the absolute or relative size or velocity direction for an item.
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What is the current moving through a wire that has 0. 60 C of electrons passing through a point in 1. 5 min?
The current moving through a wire that has 0. 60 C of electrons passing through a point in 1. 5 min is 0.4 Amps (A).
The current is 0.4 Amps (A),
It is calculated using the formula
I = Q/t,
= 0.60/1.5
= 0.4 Amps
where Q is the charge (Coulombs) of electrons and t is the time (seconds) taken for them to pass through the point.
The positive sign for current corresponds to the direction a positive charge would move. In metal wires, current is carried by negatively charged electrons, so the positive current arrow points in the opposite direction the electrons move.
In fact, an electric current is defined as the net movement of charge(s). This includes slow drift of electrons inside a wire, or a stream of electrons, protons, or muons traveling near light speed through the vacuum inside a particle accelerator.
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A man of mass 50kg ascends a flight of stairs 5m high in 5 seconds.If acceleration due to gravity is 10m/s the power expended is
Answer:
490 W
Explanation:
F = ma = (50 kg)(9.8 m/s²) = 490 N (Weight = W = mg)
Work = W = Fd = (490 N)(5 m) = 2450 J
Power = W/t = 2450 J / 5 s = 490 J/s = 490 W (watts)
If you are traveling at 50 mph, how many minutes will it take to travel 10 km?
( 1 mile = 1. 6 kilometer )
To calculate time, divide the distance by speed. hence, the time is 7.5 minutes.
Divide the distance by the speed to find the passing duration. Multiply the speed by the duration to get the distance. These formulae can be represented more simply as s=d/t, where s stands for speed, d for distance, and t for time.
Given: speed= 50 mph.
distance= 10km
1 mile = 1.6 km
1km = 1/1.6 miles= 0.625 milesmiles
We know that, time= distance/speedspeed
Time = 0.625*10/50= 0.125 hours.
1 hour = 60 minutes
hence, 0.125 hours= 60*0.125 = 7.5 minutes.
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What do we call a push or pull on an object?
Answer:
a force
Explanation:
A transverse wave is observed to be moving along a lengthy rope. Adjacent crests are positioned 2. 4 m apart. Exactly six crests are observed to move past a given point along the medium in 9. 1 seconds. Determine the wavelength, frequency and speed of these waves
The wavelength of the wave is 2.4 m, the frequency of the wave is 0.6593 Hz, and the speed of the wave is 0.2637 m/s.
A transverse wave is observed to be moving along a lengthy rope
Distance between adjacent crests (wavelength) = λ = 2.4 m
Number of crests passing a point = 6
Time taken for these crests to pass = t = 9.1 s
We can use the formula:
Speed = Distance / Time
Speed of the wave (v) = λ / t
Frequency of the wave (f) = Number of crests / Time taken
So, substituting the given values, we get:
Speed of the wave (v) = λ / t
v = 2.4 m / 9.1 s
v = 0.2637 m/s
Frequency of the wave (f) = Number of crests / Time taken
f = 6 / 9.1 s
f = 0.6593 Hz
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which graph best represents the relationship between the between the strength of an electric field and distance from a point charge?
The fourth option best represents the proper graph, which resembles a branch of a hyperbola.
What is the Coulomb's law?From Coulomb's Law.
F = k · (q₁ · q₂) / r²
where
Q1 and Q2 are the charges of the two particles, F is the electrostatic force separating them, k is the Coulomb's constant, and r is the separation between them.
The force becomes weaker as the two particles move apart.
Moreover, as r approaches zero, F approaches 1 / 0 =. F is endlessly powerful when there's no separation between the two particles. As a result, the force axis will never be touched by the force about distance graph.
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Find the period ,acceleration and velocity of a body in SHM, whose amplitude is 10cm and frequency 100Hz
i. the period of oscillation: Speed, V=ωr2−y2. ii. the acceleration at the maximum displacement: V=10π×(0.05)2−(0.03)2=10π×0.04=0.4π m/s.
iii. the velocity at the center of motion: V=10π×(0.05)2−02
given = 5 cm = 0.05 m
T = 0.2 s
ω=2π/T=2π/0.2=10πrad/s
At the point when relocation is y, then acceleration, a=−ω2y
Speed, V=ωr2−y2
Case (a) When y=5cm=0.05m
a=−(10π)2×0.05=−5π2m/s2
V=10π×(0.05)2−(0.05)2=0
Case (b) When y=3cm=0.03m
a=−(10π)2×0.03=−3π2m/s2
V=10π×(0.05)2−(0.03)2=10π×0.04=0.4π m/s
Case (c) When y=0
a=−(10π)2×0=0
V=10π×(0.05)2−02
T = 2π√(m/k). By timing the length of one complete oscillation we can decide the period and thus the frequency. Note that on account of the pendulum, the period is free of the mass, while on account of the mass of spring, the period is free of the length of the spring.
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the complete question is:
A body moving in simple harmonic motion has an amplitude of 10cm and a frequency of 100 Hz. Find i. the period of oscillation, ii. the acceleration at the maximum displacement, iii. the velocity at the center of motion.
A bullet is fired with a speed of 50 m/s into a block of wood of mass 0. 5kg and becomes embedded in it. If it gives block a speed of 15m/s. Find the mass of the bullet correct to 2sf
The mass of the bullet is approximately 0.21 kg (to 2 significant figures).
We can use the principle of conservation of momentum to solve this problem. The total amount of momentum prior to and following the impact are equal.
Let the mass of the bullet be "m" kg.
The momentum of the bullet before the collision is:
p1 = m × 50 m/s = 50m
The momentum of the bullet and block after the collision is:
p2 = (m + 0.5 kg) × 15 m/s = 15(m + 0.5)
According to the principle of conservation of momentum:
p1 = p2
50m = 15(m + 0.5)
50m = 15m + 7.5
35m = 7.5
m = 7.5/35 = 0.2143 kg (rounded to 2 significant figures)
Therefore, the mass of the bullet is approximately 0.21 kg (to 2 significant figures).
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The kinetic energy of a bullet fired from a gun is 40j. If the mass of the bullet is 0.1kg, calculate the initial speed of the bullet
Answer:
KE = 1/2 M V^2
V = (2 KE / M)^1/2 = (2 * 40 / .1)^1/2 = 28.3 m/s
An 8kg ball traveling at4m/s collides head on with a 3kg ball traveling at 14m/s. The balls bounce off each other and travel back the way they come. The 8kg ball travels away at 2m/s. Calculate the kinetic energy before and after collision
The kinetic energy before the collision is 358 J and the kinetic energy after the collision is 232 J.
KE_before = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
KE_before = (1/2) * 8 * 4^2 + (1/2) * 3 * 14^2
= 64 + 294
= 358 J
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
Substituting the given values, we get:
8 * 4 + 3 * 14 = 8 * 2 + 3 * v2'
Solving for v2', we get:
v2' = (8 * 4 + 3 * 14 - 3 * v1') / 3
The negative sign indicates that the 3kg ball is moving in the opposite direction after the collision. We also know that the 8kg ball is moving away at 2m/s. Therefore,
v1' = 2 m/s
Substituting this value, we get:
v2' = (8 * 4 + 3 * 14 - 3 * 2) / 3
= 12 m/s
The kinetic energy after the collision is:
KE_after = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
Substituting the given values, we get:
KE_after = (1/2) * 8 * 2^2 + (1/2) * 3 * 12^2
= 16 + 216
= 232 J
Kinetic energy is the energy an object possesses due to its motion. In physics, it is defined as the energy that an object possesses due to its motion, and it is dependent on the mass and velocity of the object. The formula for kinetic energy is KE = 1/2mv², where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.
Kinetic energy is a scalar quantity, meaning that it has only magnitude and no direction. It is a fundamental concept in physics, and it is used to describe many phenomena, including the motion of particles, the motion of objects, and the conversion of energy between different forms. The kinetic energy of an object can be transformed into other forms of energy, such as potential energy or thermal energy, through various processes.
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What is the minimum diameter at section (1) to avoid cavitation at that point? take d2 = 12 cm, the maximum water temperature as 30°c, and the density of water as 1000 kg/m3. (round the final answer to three decimal places. )?
The minimum diameter at section (1) to avoid cavitation at that point is 5.057 cm (rounded to three decimal places).
How to determine the minimum diameter at section (1) to avoid cavitation at that point?To determine the minimum diameter at section (1) to avoid cavitation at that point, we can use the following formula:
d1 = √((4q)/((πv)),
where
q = the flow rate (m³/s)v = the velocity of water (m/s)
To avoid cavitation, the velocity of water at section (1) should not exceed the critical velocity, which can be calculated using the following formula:
vc = √((p2-pv)/(0.5*density)),
where
p2 = the pressure at section (2) (Pa)
pv = the vapor pressure of water at the maximum temperature (Pa)
density = the density of water (kg/m³)
Assuming that the pressure at section (2) is atmospheric (101325 Pa), the vapor pressure of water at 30°C is 4243.7 Pa.
Thus,
vc = √((101325-4243.7)/(0.5*1000)) = 14.697 m/s
To determine the flow rate, we can use the continuity equation:
q = ((π/4)d2²v
where d2 is given as 12 cm, or 0.12 m. If we assume that the flow is steady and incompressible, the flow rate is constant throughout the pipe.
Thus,
q = ((π/4)(0.12)²v
Combining these equations, we get:
d1 = √((4*((π/4)(0.12)²v)/((πvc))
d1 = √((0.0144v)/14.697)
d1 = 0.003762 × [tex]v^{0.5[/tex]
To avoid cavitation, d1 must be greater than or equal to the minimum diameter required at section (1). Therefore, we can set d1 equal to the critical diameter, which is given by:
dc = √((4q)/((πvc))
Substituting the values for q and vc, we get:
dc = √((π/4)(0.12)² 14.697)
dc = 0.05057 m
Therefore, the minimum diameter at section (1) to avoid cavitation at that point is 5.057 cm (rounded to three decimal places).
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you have a camera with a 35.0-mm-focal-length lens and 36.0-mm-wide film. you wish to take a picture of a 12.0-m-long sailboat but find that the image of the boat fills only of the width of the film. (a) how far are you from the boat? (b) how much closer must the boat be to you for its image to fill the width of the film?
The distance between the boat and the camera is 0.0525 m. The boat must move closer by a distance of 35 mm to fill the width of the film.
you have a camera with a 35.0-mm-focal-length lens and 36.0-mm-wide film. you wish to take a picture of a 12.0-m-long sailboat but find that the image of the boat fills only the width of the film.
(a) Given data:
The focal length of the lens, f = 35.0 mm
Width of the film, w = 36.0 mm
Length of the sailboat, L = 12.0 m
Length of the image of the sailboat on film = Width of the film
Let the distance between the boat and the camera be x.
The object distance, u = ?
Image distance, v = f
Let M be the magnification of the image, then we have:
M = v/u
Width of the image, W' = w
Since the image of the boat fills only half of the width of the film,
W' = w/2.
Now, using the relation,
1/f = 1/u + 1/v
Putting f = 35mm, v = 35mm, and solving for u, we get
u = 52.5 mm = 0.0525 m
Therefore, the distance between the boat and the camera is 0.0525 m.
(b)Width of the image, W = w = 36 mm
Width of the sailboat, L = 12 m
Let the new distance between the boat and the camera be y.
The new image distance, v' = f
The new object distance, u' =?We know that the magnification of the image is given by M = v'/u'.
And the width of the image of the boat is given by:
W = (M)(L)
Using the relation,
1/f = 1/u' + 1/v', we can find the object distance, u'.
Substituting the given values, we get
u' = 0.0175 m = 17.5 mm
Therefore, the boat must move closer by a distance of (52.5 - 17.5) mm = 35 mm to fill the width of the film.
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A child in danger of drowning in a river is being carried downstream by a current that flows uniformly with a speed of 2.0 m/s. The child is 200 m away from the show and 1500 m upstream of the boat dock from which the rescue team sets out. If their boat speed is 8.0 m/s with respect to the water, at what angle should the pilot leave the shore to go directly to the child?
The pilot should leave the shore at an angle of 14.04° in order to go directly to the child.
To find the angle, we can use the Pythagorean Theorem and trigonometry.
First, we need to find the distance between the child and the boat dock. Using the Pythagorean Theorem, we get:
d = √((200 m)^2 + (1500 m)^2) = 1520.69 m
Next, we need to find the time it will take for the child to reach the boat dock. Since the child is being carried by the current at a speed of 2.0 m/s, we can use the formula:
t = d/v = 1520.69 m / 2.0 m/s = 760.34 s
Now, we need to find the distance the boat will travel in this time. Since the boat has a speed of 8.0 m/s with respect to the water, we can use the formula:
d = v*t = 8.0 m/s * 760.34 s = 6082.72 m
Finally, we can use trigonometry to find the angle at which the pilot should leave the shore. Using the formula:
tan(θ) = opposite/adjacent = 200 m / 6082.72 m = 0.0329
θ = tan^-1(0.0329) = 14.04°
Therefore, the pilot should leave the shore at an angle of 14.04° in order to go directly to the child.
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Drive north on a s You straight two lane road at constant 88km/h. A truck in the other lane approaches you at a constant 104km/h. Find (a) the truck's velocity relative to you and (b) your velocity relative to the truck. (c) How do the relative velocities ties changes after you and the truck pass each other?
(a) To find the truck's velocity relative to you, we need to subtract your velocity from the truck's velocity.
Since you are driving north and the truck is approaching you in the other lane, the velocities are in opposite directions. Therefore, we can simply subtract the two speeds: Truck's velocity relative to you = Truck's velocity - Your velocity = 104 km/h - 88 km/h = 16 km/h north (since the truck is approaching you). Therefore, the truck's velocity relative to you is 16 km/h north. (b) To find your velocity relative to the truck, we need to subtract the truck's velocity from your velocity. Since you are driving north and the truck is approaching you from the opposite direction, the velocities are again in opposite directions. Therefore, we can simply subtract the two speeds: Your velocity relative to the truck = Your velocity - Truck's velocity = 88 km/h - 104 km/h = -16 km/h north (since you are moving away from the truck). Therefore, your velocity relative to the truck is 16 km/h south. (c) After you and the truck pass each other, the relative velocities change. The truck is now moving away from you, and you are moving away from the truck. Therefore, the velocity of the truck relative to you will decrease, while your velocity relative to the truck will increase. However, the magnitude of the relative velocities will remain the same, since the difference between the two velocities is still 16 km/h. The direction of the relative velocities will also change, with the truck's velocity relative to you now being 16 km/h south, and your velocity relative to the truck now being 16 km/h north.
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Gino is teaching his partner a dance by using eight counts.
He has included quarter-turns at the end of each eight count for a total of 32 counts.
When Gino's dance partner finishes the dance, she has faced a total of three walls.
What does Gino need to tell his partner through constructive feedback?
she performed too many quarter-turns.
She danced for longer than 32 counts.
She used bad technique for the dance.
She forgot an eight count or a quarter-turn.
Answer:
uhhh
Explanation:
Gino should tell his partner that she should have only done four quarter turns and should have finished the dance after 32 counts. This would have resulted in her facing only two walls at the end.
What is quarter?Quarter is a term used to describe the amount of rotation an object has undergone when it does a full turn. A quarter-turn is the equivalent of 90 degrees. Quarter are commonly used to measure the rotation of a wheel, a lever, a bolt, or any other object that can be rotated. It is also used in architecture to describe the amount of rotation of a staircase or other structure. In engineering, quarter-turns are used to measure the amount of torque that is being applied to a mechanism or device. Quarter are also used to measure the amount of rotation when adjusting the settings of a device.
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shows a map of Olivia's trip to a coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8 mi to where Broadway turns, and then continues another 1.4mi to the shop.
What is the magnitude of the total displacement of her trip?
What is the direction of the total displacement of her trip?
The magnitude of the total displacement of Olivia's trip is 1.9 miles. The direction of the total displacement of her trip is southeast.
To find the magnitude of the total displacement, we need to use the Pythagorean theorem:
a^2 + b^2 = c^2
In this case, a = 0.9 miles (the distance she travels south), and b = 0.8 + 1.4 = 2.2 miles (the distance she travels east).
Plugging in these values, we get:
0.9^2 + 2.2^2 = c^2
0.81 + 4.84 = c^2
5.65 = c^2
c = √5.65
c = 2.376
So the magnitude of the total displacement is approximately 2.376 miles.
To find the direction of the total displacement, we need to use the tangent function:
tan θ = opposite/adjacent
In this case, opposite = 2.2 miles (the distance she travels east), and adjacent = 0.9 miles (the distance she travels south).
Plugging in these values, we get:
tan θ = 2.2/0.9
θ = tan^-1(2.2/0.9)
θ = 67.38 degrees
So the direction of the total displacement is approximately 67.38 degrees from south, or southeast.
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Explain the forces acting on the skateboard and how the forces affect the motion of the skateboard.
The forces acting on a skateboard include gravity, friction, air resistance, and applied force, and they all affect the motion of the skateboard in different ways.
When a skateboard is in motion, several forces act on it that affect its movement. The main forces acting on a skateboard are:
Gravity: This is a force that pulls the skateboard downwards towards the ground. The weight of the skateboard and the rider are also affected by gravity.
Friction: This is a force that opposes the motion of the skateboard. Friction is caused by the skateboard's wheels rolling on the ground or surface, and it can slow down the skateboard's speed.
Air resistance: This is a force that opposes the motion of the skateboard as it moves through the air. The faster the skateboard moves, the more air resistance it experiences.
Applied force: This is the force applied by the rider to propel the skateboard forward or to perform tricks. The amount of applied force determines how fast or slow the skateboard moves.
The combination of these forces affects the motion of the skateboard. If the applied force is greater than the forces of friction, air resistance, and gravity, the skateboard will accelerate. If the forces of friction, air resistance, and gravity are greater than the applied force, the skateboard will slow down or come to a stop.
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super mario, spying a mystery block just ahead and wanting star power, takes a running leap towards the block, which is a distance of 8.2 m above mario's head. the acceleration due to gravity in the mushroom kingdom is roughly 1/5th of that on earth, gmk
The time taken by Super Mario is 2.91 seconds to reach the mystery block.
To find the time it takes for Super Mario to reach the mystery block, we can use the kinematic equation:
∆y = v0t + (1/2)at2
Where ∆y is the change in a vertical position, v0 is the initial vertical velocity, a is the acceleration due to gravity, and t is the time.
Since the mystery block is 8.2 m above Mario's head, ∆y = 8.2 m. The initial vertical velocity, v0, is 0 m/s since Mario is starting from rest. The acceleration due to gravity in the Mushroom Kingdom, gmk, is 1/5th of that on Earth, so
a = (1/5)g = (1/5)(9.8 m/s2)
a = 1.96 m/s2.
Plugging these values into the equation, we get:
8.2 m = (0 m/s)t + (1/2)(1.96 m/s2)t2
Simplifying and rearranging, we get:
t2 = (8.2 m)/(0.98 m/s2)
Taking the square root of both sides, we get:
t = √((8.2 m)/(0.98 m/s2)) = 2.91 s
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.
After an unfortunate accident occurred at a local warehouse, you were contracted to determine the cause. A jib crane collapsed and injured a worker. The horizontal steel beam had a mass of 85. 10 kg per meter of length, and the tension in the cable was =12040 N. The crane was rated for a maximum load of 500 kg. If =5. 000 m, =0. 450 m, =2. 000 m, and ℎ=2. 250 m, what was the magnitude of L (the load on the crane) before the collapse? The acceleration due to gravity is =9. 810 m/s2
The magnitude of the load on the crane before the collapse was 7871.48 N, which is well below the maximum load rating of 500 kg.To determine the load on the crane, we need to use the principles of static equilibrium.
The crane is in equilibrium when the sum of the forces acting on it is zero and the sum of the torques is also zero.The forces acting on the crane are tension in the cable and the weight of the horizontal beam. the torque is due to the weight of the horizontal beam.First, calculate the weight of the horizontal beam:
W = mgL = 85.105.0009.810 = 4168.52 N, where m is the mass per meter of length, g is the acceleration due to gravity, and L is the length of the beam.Now calculating the torque due to the weight of the beam :
τ = W*(h/2) = 4168.52*(2.250/2) = 4686.03 Nm, where h is the height of the horizontal beam. since the crane is equilibrium, the tension in the cable must balance the weight of the beam which is T = W. now substituting the values we get :
=>12040 N = 4168.52 N + L=> 12040 N - 4168.52 N = 7871.48 N
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PLS HELP, ONLY 5 QUESTIONS !!!
1. Describe the troposphere. Be specific and provide details.
2. Describe the stratosphere. Be specific and provide details.
3. Describe the mesosphere. Be specific and provide details.
4. Describe the thermosphere. Be specific and provide details.
5. Describe the problems of the ozone layer and how it affects our environment.
1. The troposphere is the lowest layer of Earth's atmosphere. Most of the mass (about 75-80%) of the atmosphere is in the troposphere. Most types of clouds are found in the troposphere, and almost all weather occurs within this layer.
2. The stratosphere is very dry air and contains little water vapor. Because of this, few clouds are found in this layer and almost all clouds occur in the lower, more humid troposphere. Polar stratospheric clouds (PSCs) are the exception. PSCs appear in the lower stratosphere near the poles in winter.
3. The mesosphere lies between the thermosphere and the stratosphere. “Meso” means middle, and this is the highest layer of the atmosphere in which the gases are all mixed up rather than being layered by their mass. The mesosphere is 22 miles (35 kilometers) thick.
4. The thermosphere is a layer of Earth's atmosphere that is directly above the mesosphere and below the exosphere. It extends from about 90 km (56 miles) to between 500 and 1,000 km (311 to 621 miles) above our planet.
5. The ozone layer acts as a natural filter, absorbing most of the sun's burning ultraviolet ( UV ) rays. Stratospheric ozone depletion leads to an increase in UV -B that reach the earth's surface, where it can disrupt biological processes and damage a number of materials
Explanation: Hope This Helps!
b) A hammer of weight 100 N falls freely on a nail from height 1.25 m. Find the impulse and average force of blow if impact last for 10-² S.
An object is placed in front of a convex mirror at infinity. According to the New Cartesian Sign Convention, the sign of the focal length and the sign of the image distance in this case are respectively:
(a) +, -
(b) -,+
(c) -, -
(d) +,+
Answer:
the correct answer is (c) -, -.
Explanation:
When an object is placed in front of a convex mirror at infinity, the image formed is a virtual image located at the focal point of the mirror.
According to the New Cartesian Sign Convention, the focal length of a convex mirror is negative. Therefore, the sign of the focal length in this case is "-".
Since the image is formed behind the mirror and is a virtual image, its distance from the mirror is negative. Therefore, the sign of the image distance in this case is also "-".
Therefore, the correct answer is (c) -, -.
Determine the voltage dropped across
The voltage dropped across R₃ is 52.68 volts.
How to calculate voltage?To determine the voltage dropped across R₃, we can use the voltage divider formula:
V₃ = ( R₃ / ( R₁ + R₂ + R₃ + R₄ ) ) x ET
Substituting the given values, it can be calculated as:
V₃ = (36 Ω / ( 15 Ω + 18 Ω + 36 Ω + 13 Ω ) ) x 120 V
V₃ = ( 36 Ω / 82 Ω ) x 120 V
V₃ = 52.68 V
Therefore, the voltage dropped across R₃ is approximately 52.68 volts.
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A concave mirror has a focal length of 5.0 cm. A candle is located at a distance of 8.0 cm from the mirror. Calculate the image distance.
Answer:
13 cm
Explanation:
13 cm is the image distance
1. What represents the building blocks for all living things?
Cells represents the building blocks for all living things.
Cells are called "the building blocks of life" because they are the basic structural and functional elements of human life. Cells are the basic units of all stages of biological organization. This suggests that cells of different species create tissues, organs, and organ systems.
The cornerstone of all living beings is the cell. There are billions of cells in the human body, each serving a specific purpose.
Your DNA contains information that tells our cells how to make proteins. Vital bodily processes including digestion, cell growth and muscle movement are fueled by protein.
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Yellow,,white,,absorbs,,reflects ,,wavelengths. The color of an object has to do with the light ________ That are reflected from the object. A white t-shirt looks _______ because it _____ all the colors while black ______ all the wavelengths of light. A ripe mango is yellow because when white light hits it ,its atomic imards reflect all the _________ wavelengths more than the other colors and bounce them at our eyes.
The color of an object has to do with the light wavelength. A white t-shirt looks yellow because it absorbs all the colors while black white all the wavelengths of light. A ripe mango is yellow because reflect all the reflects wavelengths.
The light wavelength has an impact on an object's color of which the object reflects. Because it absorbs all hues, a white t-shirt seems yellow, but a black one emits all light wavelengths. A ripe mango appears yellow because, when white light strikes it, its atomic impurities bounce all the wavelengths of light at our sight more than the other colors.
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