Given:
The mass of the child and the sled is,
[tex]m=25.2\text{ kg}[/tex]The child acquires a speed at the bottom of the hill is,
[tex]v=6\text{ m/s}[/tex]To find:
The height of the hill
Explanation:
The potential energy at the top of the hill will be converted to the potential energy at the bottom of the hill. So for the hill of height 'h', we can write, the potential energy as
[tex]PE=mgh[/tex]The kinetic energy is
[tex]KE=\frac{1}{2}mv^2[/tex]So,
[tex]\begin{gathered} mgh=\frac{1}{2}mv^2 \\ h=\frac{v^2}{2g} \\ h=\frac{6^2}{2\times9.8} \\ h=1.84\text{ m} \end{gathered}[/tex]Hence, the height of the hill is 1.84 m.
An unwary football player collides head-on with a padded goalpost while running at 7.5 m/s and comes to a full stop after compressing the padding and his body by 0.27 m. Take the direction of the player’s initial velocity as positive.1.assuming constant acceleration calculate the his acceleration during the collision in meters per second squared.2 how long does the collision last in seconds.
Answers:
1. a = -104.16 m/s²
2. t = 0.072
Explanation:
To find the acceleration, we will use the following equation:
[tex]v^2_f=v^2_i+2ax[/tex]where vf is the final velocity, vi is the initial velocity, a is the acceleration and x is the distance. So, replacing vf by 0 m/s, vi by 7.5 m/s, and x by 0.27m, we get:
[tex]\begin{gathered} 0^2=7.5^2+2a(0.27) \\ 0=56.25+0.54a \end{gathered}[/tex]Then, solving for a, we get:
[tex]\begin{gathered} 0-56.25=56.25+0.54a-56.25 \\ -56.25=0.54a \\ \frac{-56.25}{0.54}=\frac{0.54a}{0.54} \\ -104.16m/s^2=a \end{gathered}[/tex]Therefore, the acceleration during the collision is -104.16 m/s²
Then, to calculate how long the collision last, we will use the following equation:
[tex]v_f=v_i+at[/tex]So, replacing the values and solving for t, we get:
[tex]\begin{gathered} 0=7.5-104.16t \\ 104.15t=7.5 \\ t=\frac{7.5}{104.15}=0.072s \end{gathered}[/tex]Therefore, the collision last 0.072 seconds
A 6.5 kg lump of clay is sliding to the right on a fricitonless surface with a speed of 23 m/s. It collides head-on and sticks to a 2 kg metal sphere that is sliding to the left with a speed of -7 m/s. What is the kinetic energy of the combined objects after the collision?
The kinetic energy of the combined objects after the collision = 1768.25 Joules
Explanation:The mass of the lump of clay, m₁ = 6.5 kg
The speed of the lump of clay, v₁ = 23 m/s
The mass of the metal sphere, m₂ = 2 kg
The speed of the metal sphere, v₂ = -7 m/s
The Kinetic Energy (KE) of the combined objects after collision is calculated as shown below:
[tex]KE=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2^{}[/tex][tex]\begin{gathered} KE=\frac{1}{2}(6.5)(23^2)+\frac{1}{2}(2)(-7)^2 \\ KE=1719.25+49 \\ KE\text{ = }1768.25J \end{gathered}[/tex]The kinetic energy of the combined objects after the collision = 1768.25 Joules
The acceleration of an object is ___proportional to the net force and ___ proportional to its mass? Can you tell me which ones it would be from these options?directly, directlydirectly, inverselyinversely, inverselyinversely, directly
Answer:
the acceleration of an object is directly proportional to the net force and inversely proportional to its mass.
Explanation:
when the net force on an object goes up, so does the acceleration, meaning one increases as the other increases
the acceleration of an object decreases as the mass increases, therefore this relationship is inversely proportional
A 6 kg object is being pulled by a horizontal force F=120 N on a friction-less horizontal surface. It moved a distance of 18 m. If its initial kinetic energy was 100 Joules, what is the final kinetic energy in Joules?
The final kinetic energy is 120m.
What is Work-Energy Theorem?
The Work-Energy Theorem states that the work done is equal to the change in the K.E. i.e Kinetic Energy of the object.
W = Δ(K.E.)
In the given question we had,
Mass = 6 kg,
Force = 120 N,
Distance = 18 m,
Initial Kinetic Energy ( KE1 ) = 100 Joules
According to Work-Energy Theorem,
W = Δ(K.E.)
W = KE2-KE1
F x S = KE2 - 100
120 x 18 = KE2 - 100
2160 + 100 = KE2
2260J = KE2
So, the final kinetic energy is 2260J.
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If the mass of the vase with flowers is 3.2 kg, what is the magnitude of the normal force?
The magnitude of normal force will be 31.36 N but in the opposite direction of the vase's weight.
We know that forces exist in pairs.
According to Newton's third law of motion,
" Every action has an equal and opposite reaction "
Also,
The second law of motion states, "The magnitude of a force is equal to the product of mass and acceleration acting upon it."
So, keeping in mind the above two laws:
The weight of the vase will be equal to:
W = mass of the vase × acceleration due to gravity
W = m × g
W = 3.2 kg × 9.8 m/s²
W = 31.36 N
So, the magnitude of normal will be the same as the Weight but the direction of Normal will be opposite to the direction of the weight.
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REWRITE THIS SENTENCE ON THE LINES PROVIDED.
The time it takes for the moon to rotate once on its axis is the same amount of time it takes for
the moon to revolve around the Earth once. The result is that the SAME SIDE OF THE MOON
ALWAYS FACES THE EARTH.
The amount of time it takes for the moon to complete one full rotation around the Earth is the same as its period of one full rotation on its axis. Due to this, the EARTH ALWAYS SEE THE SAME SIDE OF THE MOON.
What does the earth's rotation entail?With an inclination of 23.45 degrees from the plane of its orbit around the sun, the Earth revolves on its axis in relation to the sun every 24.0 hours mean solar time. The differences brought on by the Earth's are averaged out to create mean solar time.
What occurs when the Earth rotates?Rotation causes the day-night cycle, which in turn induces a cycle of temperature and humidity. As the world spins, the sea level rises and falls twice each day. The tidal range is determined by the gravitational pull of the sun and moon together
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state the dimension of energy in physics
hint: Energy = force × distance
Force = Mass * Acceleration = kg* m/s^2= MLT^-2
Distance = metres= L
Energy = MLT^-2 * L =ML^2T^-2
which are neutrally charged, are found in thenucleus of the atom.
Given:
Nucleus of the atom
Required:
Neutra
a student drops a pebble from the edge of a vertical cliff. the pebble hits the ground 4 s after it was dropped. what is the height of the cliff? a. 20 m b. 40 m c. 60 m d. 80 m
The object's speed shortly before it lands on the earth is 40 m/s.
What is an example of velocity?The speed at which something moves in a specific direction is known as its velocity. as the speed of a car driving north on a highway or the pace at which a rocket takes off. Because the velocity vector is scalar, its absolute value magnitude will always equal the motion's speed.
The parameters are as follows: the pebble's time, t = 4 s; the object's velocity right before impact;
The kinematic equation is as follows;
v = in which
v = 0+10 (4)
The object's speed right before impact with the earth is v = 40 m/s2, where g is the acceleration caused by gravity and an is a constant of 10 m/s2. As a result,
the object's final velocity before impact is 40 m/s.
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A NASA probe is moving horizontally above the surface of the moon at a constant speed to the right, as depicted in the diagram below. It releases an instrument package when it is directly above Point P. As seen from the lunar surface, which path would the package likely follow after the release and why?
B, because the gravity of the moon will pull the instrument to the ground with constant acceleration and the lack of an atmosphere allows the package to fall straight down
D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant
C, because the reduced gravity of the moon pulls the package down vertically at a constant speed while the package travels horizontally at a constant speed resulting in a straight-line trajectory to the lunar surface.
A, because the gravity of the moon pulls the package down with constant acceleration, while the atmosphere of the moon creates horizontal drag on the package which reduces the horizontal component of the package’s velocity causing the package to be pulled backward as it falls
As seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant a, so that its vertical velocity increases.
A is incorrect because there is no wind pushing the package backwards. B is incorrect because the package has an initial velocity. C is incorrect because vertical velocity is not constant due to the presence of gravity. E and F are incorrect because gravity acts immediately after the package is dropped.
D is correct because the horizontal component remains constant because there is no horizontal force acting on the package. This is because in outer space there is no atmosphere, so there will be no air resistance. The vertical component increases with respect to time because of constant acceleration due to gravitational pull on the package.
Therefore, as seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant
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A computer connected to a motion sensor creates velocity—time graph for a ball rolling down an incline. On the graph, the velocity increases by 0.3 m/s for every tenth of a second increment on the graph. What is the acceleration of the object? A. 0.3 m/s2. B. 1 m/s2. C. 3 m/s2. D. The acceleration cannot be determined from the given information.
We are asked to determine the acceleration of an object which speed changes by 0.3 m/s every 1/10 seconds. The definition of acceleration is the following:
[tex]a=\frac{\Delta v}{\Delta t}[/tex]Where:
[tex]\begin{gathered} \Delta v=\text{ change in velocity} \\ \Delta t=\text{ change in time} \end{gathered}[/tex]Substituting the given values we get:
[tex]a=\frac{0.3\frac{m}{s}}{\frac{1}{10}s}[/tex]Solving the operations we get:
[tex]a=3\frac{m}{s^2}[/tex]Therefore, the acceleration is option C.
Imagine that you are on board a ship that was struck by a rogue wave.
Tell your story, from the calm before the wave hit to its aftermath.
The possibility of a wave toppling cruise liner is extremely low. They are made wide and have enough ballast on the lower decks to be heavy enough rogue waves. On the side, the crew's neglect also be necessary.
What are rogue waves?A wave that is double the region's major wave height is typically considered a rogue wave. The highest one-third of waves on average over a period of time make up the noteworthy wave height. Even the biggest ships and oil rigs can be rendered useless and sunk by rogue waves.
Have rogue waves ever struck a cruise ship?Rogue waves have occasionally hit cruise ships, although it is not frequently. Four cruise ships have collided with rogue waves since rogue wave records were first kept in 1995. All sustained damage, and some people reported injuries, but there have been no confirmed fatalities on cruise ships due to rogue waves.
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(20%) Problem 5: Two identical springs, A and B, each with spring constant k = 54.5 N/m, support an object with a weight W = 11.6 N. Each spring makes an angle of 0 = 20.6 degrees to the vertical, as shown in the diagram. Create an expression for the tension in spring A
The tension in spring A is T = W/(2cosθ)
What is tension?Tension is the stretching force in a spring.
How to find the expression for the tension in the spring?Let
T = the tension in the each spring, W = weight andθ = angle each spring makes with the verticalResolving the tension in each spring vertically, so we can have that
for spring A, the tension is Tcosθ and for spring B, the tension is TcosθNow the vertical component of the tension in each equals the weight. So, we have that
Tcosθ + Tcosθ = W
Adding them together, we have that
2Tcosθ = W
Dividing both sides by 2cosθ, so, we can have that
T = W/2cosθ
Thus, the tension in each spring is T = W/(2cosθ)
So, the tension in spring A is T = W/(2cosθ)
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A heart defibrillator passes 10.3 A through a patient's torso for 5.00 ms in an attempt to restore normal beating.(a) How much charge passed?(b) What voltage was applied if 492 J of energy was dissipated?KV(c) What was the path's resistance?ΚΩ(d) Find the temperature increase caused in the 8.00 kg of affected tissue. The specific heat of tissue is 3500 J/(kg. °C).°C
a) The formula for calculating the quantity of charge is expressed as
Q = IT
where
Q is the quantity of charge
I is the current
T is the time
From the information given,
I = 10.3
T = 5 ms = 5 x 10^-3 s
Q = 10.3 x 5 x 10^-3
Q = 51.5 x 10^- 5 C
The quantity of charge passed is 51.5 x 10^- 5 C
b) The formula for calculating the energy is expressed as
E = I^2RT
where
R is the resistance
E is the energy
From the information given,
E = 492 J
Thus,
492 = 10.3^2 x R x 5 x 10^-3
R = 492/(5 x 10^-3 x 10.3^2)
R = 927.514 ohms
Voltage, V = IR
Voltage = 10.3 x 927.514
Voltage = 9553.398 V
We would divide by 1000. It becomes
Voltage = 9.553 KV
c) From the calculations,
Resistance = 927.514 ohms
We would divide by 1000. It becomes
Resistance = 0.93 ΚΩ
d) Let the temperature increase be t
mass of tissue, m = 8 kg
Specific heat of tissue = 3500 J/(kg. °C).
°C
The formula for calculating the quantity of heat is
H = mcθ
where
H is the quantity of heat
From the informtaion given,
H = 492
θ = t
Thus,
492 = 8 x 3500 x t
t = 492/(8 x 3500)
t = 0.018
The temperature increase is 0.018 degrees
A 10.0 cm tall object is placed 6.00 cm in front of a curved mirror and produces an image 2.00 cm behind the mirror. What is the focal length of the mirror?0.667 cm1.50 cm-3.00 cm-0.333 cm
Given data:
The height of object is h₀=10.0 cm.
The object distance is u=6 cm.
The image distance is v=-2.00 cm.(negative because the image is behind the mirror)
The focal length can be calculated by the mirror's formula as,
[tex]\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \frac{1}{f}=\frac{1}{6}+\frac{1}{-2} \\ f=-3.00\text{ cm} \end{gathered}[/tex]Thus, the focal length of the mirror is -3.00 cm.
Two long parallel wires 0.552 meter apart are each carrying 1.75 amperes of current, as shown(a) Find the magnitude and direction of the magnetic field at point A due to the current in the top wire. (b) Find the magnitude and direction of the force per unit length this field exerts on the bottom wire.
,Given,
Distance between two-wire, d=0.552 m
Current through the wires, I=1.75 A
(a) The magnetic field at a point on the bottom wire due to top wire is given by,
[tex]B_a=\frac{\mu_0I_1}{2\pi d}[/tex]Where μ₀ is the permeability of the free space.
The direction is given by the right-hand thumb rule. According to this, the direction of the magnetic field produced by top wire will into the plane of the two wires.
On substituting the known values in the above equation,
[tex]B_a=\frac{4\pi\times10^{-7}\times1.75}{2\pi\times0.552}=6.34\times10^{-7}\text{ T}[/tex]The force per unit length is given by,
[tex]F=I_2\times B_a[/tex]The direction is given by the right-hand rule. According to this the force is directed towards the top wire.
On substituting the known values in the above equation,
[tex]F=1.75\times6.34\times10^{-7}=1.11\times10^{-6}\text{ N}[/tex]Therefore the magnetic field acting on the bottom wire due to the current in the top wire is 6.34×10⁻⁷ T and the magnetic force due to this field is 1.11×10⁻⁶ N
Moving a charge from point A, where the potential is 369.53 V, to point B, where the potential is 217.85 V, takes 0.55 milliJ of work. What is the value of the charge, in micro-Coulombs?
Given:
The potential of point A is
[tex]V_A=369.53\text{ V}[/tex]The potential of point B is
[tex]V_B=\text{ 217.85 V}[/tex]The work done is W = 0.55 milli Joule.
To find the value of charge in micro Coulomb.
Explanation:
The charge can be calculated as
[tex]\begin{gathered} Q=\frac{W}{V_B-V_A} \\ =\frac{0.55\times10^3\text{ J}}{217.85-369.53} \\ =-3.63\text{ C} \\ =\text{ -3.63 C}\times\frac{10^6\text{ }\mu C}{1\text{ C}} \\ =-3.63\times10^6\text{ }\mu C\text{ } \end{gathered}[/tex]I got first part correct but dont know how to solve second part: Two new particles with identical positive charge 3 are placed the same 0.0809 m apart. The force between them is measured to be the same as that between the original particles. What is 3 ?
Answer:
5.92 *10^-6 C
Explanation:
For the two charges q3 the force between them is given by
[tex]F=k\frac{q_3\times q_3}{d^2}[/tex]Now we know that
F = 48.1 N, d = 0.0809 m, and k = 8.99 *10^9 kg⋅m^3⋅s^−2⋅C^-2; therefore, the above gives
[tex]48.1=(8.99\times10^9)\frac{q_3\times q_3}{(0.0809)^2}[/tex][tex]\Rightarrow48.1=(8.99\times10^9)\frac{(q_3)^2}{(0.0809)^2}[/tex]Now we solve for q_3.
Dividing both sides by 8.99 * 10^9 gives
[tex]\frac{48.1}{(8.99\times10^9)}=\frac{(q_3)^2}{(0.0809)^2}[/tex]multiplying both sides by (0.0809)^2 gives
[tex]\frac{48.1}{(8.99\times10^9)}\times\mleft(0.0809\mright)^2=(q_3)^2[/tex]finally, taking the square root of both sides gives
[tex]\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}=\sqrt{(q_3)^2}[/tex][tex]q_3=\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}[/tex]Evaluating the right-hand side gives
[tex]\boxed{q_3=_{}5.92\times10^{-6}C\text{.}}[/tex]Hence, the charge q_3 is 5.92 x 10^-6 C.
Whats the percent of 10 of 20
In order to determine the associated percent of 10 related to 20, proceed as follow:
If x is the percentage, then, you can write:
[tex]\frac{x}{100}\cdot20=10[/tex]which means that x percentage of 20 is equal to 10. By solving for x, you get:
[tex]\begin{gathered} x=\frac{10}{20}\cdot100 \\ x=50 \end{gathered}[/tex]Hence, 10 is the 50% of 20
Answer:2
Explanation:multiply 0.20 times 10 you get it
In a lightning discharge, 45 C of charge move through a potential difference of 1.0 x108 V in 0.030 s.A. What is the current of the lightning strike?B. How much energy is released by the lightning bolt?
Given:
Charge, Q = 45 C
Potential difference, V = 1.0 x 10⁸ V
Time, t = 0.030 s
Let's solve for the following:
• (A). What is the current of the lightning strike?
To find the current, apply the formula:
[tex]I=\frac{Q}{t}[/tex]Where:
I si the current
Q is the charge = 45 C
t is the time = 0.030 s
Thus, we have:
[tex]\begin{gathered} I=\frac{45}{0.030} \\ \\ I=1500\text{ A} \end{gathered}[/tex]Therefore, the current of the lightning strike is 1500 Amperes.
• (B). How much energy is released by the lightning bolt?
To find the amount of energy released, apply the formula:
[tex]E=V\times Q[/tex]where:
E is the Energy released
V is the potential difference, V = 1.0 x 10⁸ V
Q is the charge = 45 C
Thus, we have:
[tex]\begin{gathered} E=1.0\times10^8\ast45 \\ \\ E=4.5\times10^9\text{ J} \end{gathered}[/tex]Therefore, the energy released is 4.5 x 10⁹ Joules.
ANSWER:
(a). 1500 A
(b). 4.5 x 10⁹ J
Tritium has a half-life of 12.3 years. How many years will have elapsed when the radioactivity of a tritium sample has decreased to 10 percent of its original value?
Given
Half life of tritium is
[tex]t_{\frac{1}{2}}=12.3\text{ years}[/tex]The sample is reduced to 10% of its original value.
To find
How many years will have elapsed when the radioactivity of a tritium sample has decreased to 10 percent of its original value?
Explanation
The activity is given by
[tex]\begin{gathered} A=A_oe^{-\lambda t} \\ \Rightarrow0.1A_o=A_oe^{-\frac{0.693}{t_{\frac{1}{2}}}t} \\ \Rightarrow ln(0.1)=-\frac{0.693}{1.23}t \\ \Rightarrow-2.302=-\frac{0.693}{1.23}t \\ \Rightarrow t=4.08 \end{gathered}[/tex]Conclusion
The time taken is 4.08 years
Three vectors are shown in this figure. Their respective moduli are A = 4.00m.B = 3, 20m and C = 2.70mCalculate 2.00 A - B + 1.30 CExpress your answer according toa) Unit vectorsb) The modulus and orientation with respect to the positive part of the x-axis
Given that,
Modulus of vector A=4.00
The angle made by the vector A with the y axis, θ₁=33.0°
The modulus of vector B=3.20 m
The angle made by the vector B with the x-axis is θ₂=40.0+90.0=130°
The modulus of the vector C=2.70 m
The angle made by the vector C with x-axis θ₃=-90°
The x and y components of the vector can be written as
[tex]\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}[/tex]Where r is the magnitude (or modulus) of the vector and θ is the angle made by the vector.
Or a vector, in cartesian coordinates, can be written as,
[tex]R=r\cos \theta\hat{\text{i}}+r\sin \theta\hat{j}[/tex]Therefore, vector A is cartesian coordinates is
[tex]\begin{gathered} \vec{A}=4.00\cos 33^{\circ}\hat{i}+4.00\sin 33.0^{\circ}\hat{j} \\ =3.35\hat{i}+2.18\hat{j} \end{gathered}[/tex]And the vector B is
[tex]\begin{gathered} \vec{B}=3.20\cos (130^{\circ})\hat{i}+3.20\sin (130^{\circ})\hat{j} \\ =-2.06\hat{i}+2.45\hat{j} \end{gathered}[/tex]And vector C is given by,
[tex]\begin{gathered} \vec{C}=2.70\cos (-90^{\circ})\hat{i}+2.70\sin (-90^{\circ})\hat{j} \\ =-2.7\hat{j} \end{gathered}[/tex]The given equation is
[tex]2.00\vec{A}-\vec{B}+1.30\vec{C}[/tex]Let this represents a vector V
On substituting the known values,
[tex]\begin{gathered} \vec{V}=2.00\vec{A}-\vec{B}+1.30\vec{C} \\ =2.00\times(3.35\hat{i}+2.18\hat{j})-(-2.06\hat{i}+2.45\hat{j)}+1.30(-2.7\hat{j}) \\ =8.76\hat{i}-1.6\hat{j} \end{gathered}[/tex](a) This is the representation with the unit vectors, where i and j are the unit vectors along the x-axis and y-axis respectively.
[tex]\vec{V}=8.76\hat{i}-1.6\hat{j}[/tex]b) The modulus of any vector is the square root of the sum of the squares of its components.
That is, the magnitude of the vector V is
[tex]\begin{gathered} V=\sqrt[]{8.76^2+(-1.6)^2} \\ =8.90\text{ m} \end{gathered}[/tex]The angle of this vector with the x-axis is given by
[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{-1.6}{8.75}) \\ =-10.36^{\circ}^{} \end{gathered}[/tex]The negative sign indicates that the vector is below the positive x-axis
Therefore the modulus of the resultant of the above equation is 8.90 m and its angle with the positive x-axis is -10.36°
What is the resistance in a circuit that has a current of 2.5A and a voltage of 40v
16 ohms
Explanation
Ohm's law relates the strength of a direct current is directly proportional to the potential difference and inversely proportional to the resistance of the circuit,it is given by the expresssion
[tex]\begin{gathered} V=IR \\ if\text{ we isolate R} \\ R=\frac{V}{I} \end{gathered}[/tex]then
Step 1
a) Let
[tex]\begin{gathered} I=2.5\text{ A} \\ V=40\text{ V} \end{gathered}[/tex]b)replace in the formula
[tex]\begin{gathered} R=\frac{V}{I} \\ R=\frac{40V}{2.5A} \\ R=16\text{ ohms} \end{gathered}[/tex]therefore, the answer is 16 ohm
I hope this helps you
Calculate the total capacitance of three capacitors 30µF, 20µF & 12µF connected in parallel across a d.c supply The answer is :Consider that the equivalent capacitance of three capacitors C1, C2 and C3 in parallel is given by:C=C1+C2+C3In this case:C1 = 30µFC2 = 20µFC3 = 12µFReplace the previous values into the formula for C and simplify:C=30μF+20μF+12μF=62μFHence, the total capacitance is 62µFQuestion 5 : Calculate the total charge on the capacitors connected in parallel if the supply voltage is 500V. Sketch a circuit diagram and label this to show how the charges are located
The circuit diagram is shown below:
From the diagram we notice that the same voltage will flow in every capacitor, this will be helpful later.
We know that this three capacitors are equivalent to a single equivalent capacitor with 62µF capacitance. The charge in this equivalent capacitor is:
[tex]Q_{eq}=(62\times10^{-6})(500)=0.031[/tex]Now, as we mentioned, the voltage is the same in each capacitor then the charge in each of them is:
[tex]\begin{gathered} Q_1=(30\times10^{-6})(500)=0.015 \\ Q_2=(20\times10^{-6})(500)=0.01 \\ Q_3=(12\times10^{-6})(500)=0.006 \end{gathered}[/tex]To check if this is correct we need to remember that the charge in the equivalent capacitor is equal to the sum of the charge in each capactior; for this case this conditon is fulfil; therefore we conclude that:
• The charge in the first capacitor is 0.015 C
,• The charge in the second capacitor is 0.01 C
,• The charge in the third capacitor is 0.006 C
The diagram with the labels is shown below:
A plane is traveling with a velocity of 70 miles/hr with a direction angle of 24 degrees. The wind is blowing at 25 miles/hr with a direction angle of 190 degrees. What is the vertical component of the wind velocity? Round your answer to the nearest whole number.
Wind velocity:
25 m/h with a direction angle of 190°.
Vertical component:
25 sin 190 = -4.34 m/s = - 4 m/s
A baseball player pitches a fastball toward home plate at a speed of 41.0 m/s. The batter swings, connects with the ball of mass 195 g, and hits it so that the ball leaves the bat with a speed of 37.0 m/s. Assume that the ball is moving horizontally just before and just after the collision with the bat.A. What is the impulse delivered to the ball by the bat? Enter a positive value if the impulse is in the direction the bat pushes the ball and enter a negative value if the impulse is in the opposite direction the bat pushes the ball. (kg m/s)B. If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat? (kN)
Given:
Initial velocity, vi = 41.0 m/s
Mass of ball, m = 195 g = 0.195 kg
Final velocity, vf = 37.0 m/s
Assuming the ball is moving horizontally just before and after collision with the bat, let's solve for the following:
• (A). What is the impulse delivered to the ball by the bat?
To find the impulse, apply the change in momentum formula:
[tex]\Delta p=p_f-p_i[/tex]Where:
pi is the initial momentum = -mvi
pf is the final momentum = mvf
Thus, we have:
[tex]\begin{gathered} \Delta p=mv_f-(-mv_i) \\ \\ \Delta p=mv_f+mv_i \\ \\ \Delta p=m(v_f+v_i) \\ \\ \Delta p=0.195(37.0+41.0) \\ \\ \Delta p=15.21\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Impulse can be said to equal change in momentum.
Therefore, the impulse delivered to the ball by the bat is 15.21 kg.m/s away from the bat.
• (B). If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat?
Apply the formula:
[tex]\text{ Impulse = Force }\ast\text{ time}[/tex]Rewrite the formula for force:
[tex]\text{ Force=}\frac{impulse}{time}[/tex]Where:
time = 3.00 m/s
impulse = 15.21 kg.m/s
Hence, we have:
[tex]\begin{gathered} \text{ F=}\frac{15.21}{3} \\ \\ F\text{ = 5.07 kN} \end{gathered}[/tex]Therefore, the magnitude of the average force exerted on the ball by the bat is 5.07 kN away from the bat.
ANSWER:
(A). 15.21 kg.m/s away from the bat
(B). 5.07 kN.
The crazed physic's student's lab partner decides to throw another pumpkin off the
roof with an initial velocity of 19.4 m/s. What is the velocity when the pumpkin
strikes the ground if it takes 3.2 seconds for it to fall?
Please Help :(
The velocity when the pumpkin strikes the ground if it takes 3.2 seconds for it to fall: 50.78 m/s
From the definition of velocity, we can find the velocity of a falling object is:
v = v₀ + gt
Here
v₀ - 19.4 m/s , t - 3.2 seconds , g - 9.80665 m/[tex]s^2\\[/tex]
v = 19.4 + (9.86 x 3.2)
v = 50.78 m/s
What is velocity?
Velocity and speed describe how quickly or slowly an object is moving. We frequently encounter circumstances when we must determine which of two or more moving objects is going faster. If the two are travelling on the same route in the same direction, it is simple to determine which is quicker. It is challenging to identify who is moving the fastest when their motion is in the other direction. The concept of velocity is useful in these circumstances. Learn about the definition of velocity in this article as well as the distinction between speed and velocity.
To learn more about velocity, refer;
https://brainly.com/question/18084516
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Ahmed wants to measure the weight of an object. What instrument should he use?
Answer:
a scale
Explanation:
you use it to weigh things
Answer:
its not a scale
Explanation:
pls brainliest
Sports biomechanics looks at athletes to better understand how the body:
A. grows.
B. creates energy.
C. moves.
D. resists disease.
Answer:
The ans is C
Explanation:
In a sporting context, biomechanics examines an athlete in relation to his or her environment and equipment. Forces acting on the body (kinetics) and movements of the body (kinematics) are analysed prior to mapping out an exercise, performance or recovery plan.
What sequence of two displacements moves from (5, 5) m to (- 5, - 5) * m while traveling a distance of exactly 20 meters? How does this distance compare to the single displacement that connects the same starting and ending point?
The two displacements that move from (5,5) to (-5,-5)
(5,5) → ( 5,-5) [10 units down]
(5,-5) → (-5,5) [ 10 units left ]
The single displacement that connects the 2 points is the hypotenuse of the formed triangle where each side is 10 m long.
Apply Pythagorean theorem:
c2 = 10^2+10^2
c^2 = 100 + 100
c^2 = 200
c =√200
c= 14.14
Compared to the simple displacement (14.14) that connects both points, it is greater.
20m > 14.14 m