When the planet is closest to the Sun, speed v and kinetic energy are the highest, and gravitational potential energy is the lowest. When the planet moves farther away, the speed and kinetic energy decrease, and the gravitational potential energy increases.
Which statement accurately describes how the acceleration of an object in free fall changes?
O A. It accelerates downward at a constant rate.
O
B. It accelerates downward at an increasing rate.
C. It accelerates downward at an irregular rate.
D. It accelerates downward at a decreasing rate.
Answer: A
Explanation: An object in free fall only is affected by gravity, so the acceleration is 9.8 m/s²
2.
A swimmer swims 1 m/s due north against a current of 3 m/s due south. What is the resultant velocity of
the boat?
The resultant velocity of the boat is equal to 2 m/s due south.
Given the following data:
Swimmer's speed = 1 m/s due northOcean current speed = 3 m/s due south.To determine the resultant velocity of the boat:
Resultant velocity can be defined as the sum (addition) of each of the vector velocity acting on a physical or an object. Thus, it is simply a combination of two or more single vector velocity.
Note: When the velocities are acting in the same direction, you will add them up while you will subtract when the velocities are acting in opposite directions.
In this scenario, the velocities are acting in opposite directions. Thus, we would subtract as follows:
[tex]V = 3 - 1[/tex]
V = 2 m/s due south.
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What is the mass of an object that has a weight of 110N ?
Answer:
The mass of the object is 11.2 Kg.
Hope you could get an idea from here.
Doubt clarification - use comment section.
The mass of an object that has a weight of 110 N is 11.2kg.
What is Weight?The weight of an object is defined as the force that acts on the object due to gravity. It is a vector quantity in some cases such as when the force of gravity acts on an object but it is also a scalar quantity where the force of gravity has a magnitude.
Weight is the gravitational force of attraction on an object due to the presence of another heavy object, such as the Earth or the Moon. It is expressed by multiplying the mass by the gravitational acceleration.
In above case,
Weight (W) = 110 N
Acceleration due to gravity (g) = 9.8 m/s^2
Let the mass of the object be m.
By using the formula,
W = mg,
[tex]110 N = 9.8 m/s^2 *m\\m = 110 N/ 9.8 m/s^2[/tex]
so, m = 11.2 Kg
Thus, the mass of an object that has a weight of 110 N is 11.2kg.
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why does polishing the surface of a metal extend fatigue life
Answer:
they are machined with shape characteristics which maximize the fatigue life of a metal.
Explanation:
they are highly polished to provide the surface characteristics which enable the best fatigue life.
what voltage is measured across the 15 ohm resistor
A single covalent bond is stronger than a single hydrogen bond so why does a group of polar molecules tend to have a higher boiling point than a group of non polar molecules
Answer:
this question makes no sense
Explanation:
like how do you get this question
Answer:They require more energy to break intermolecular forces hence polar molecules have higher melting points and boiling points than non-polar molecules of similar size, shape and number of electrons.
Explanation:
How much energy has 4x 1010 m³ of water collected in a reservoir at a 2. 3. height of 100 m from the power house? What kind of energy is that? (Given, mass of 1 m³ of water = 1000 kg)
Explanation:
[tex] \rule{999pt}{66646pt}[/tex]
Liquid X of volume 0.5m3 and density 900kgm-3 was mixed with liquid Y of volume 0.4m3 and density 800kgm-3. What was the density of the mixture?
Answer:
Density of the mixture = 855.56kgm-3
Explanation:
Density = Mass / Volume
Volume of Liquid X = 0.5m³
Density of Liquid X = 900kgm-3
Mass of Liquid X = Density × Volume
= 900kgm-3 × 0.5m³ = 450kg
Volume of Liquid Y = 0.4m³
Density of Liquid Y = 800kgm-3
Mass of Liquid Y = Density × Volume
= 800kgm-3 × 0.4m³= 320kg
As X and Y are mixed, we add their masses and volumes together:
Mass = 770kg
Volume = 0.9m³
Now we can find the density of the mixture:
Density = 770kg / 0.9m³ = 855.56kgm-3 (rounded to the 2nd decimal)
Suppose that the Sun started shrinking in size, without losing any mass. What would be the effect of the Sun's change on the orbits of the planets
Answer:
F = G M m / R^2 gravitational force on planet of mass m.
None of these quantities change in the given hypothesis so
there will be no change in the orbit of mass m
if a 5kg ball is traveling at 20 m/s and it’s stopped in 4s, what’s the impulse on the ball?
Which electromagnetic wave has a lowest frequency?
Group of answer choices
A) x-rays
B) ultraviolet light
C) microwaves
D) infrared light
E) visible light
Answer:
E.visible lights
Explanation:
hope its attachments
how to solve for time given distance and velocity
Answer:
Well, I think you're talking about kinematics, especially uniform rectilinear motion. We know that there is a specific equation for that:
S = Vt + S0
With S being the distance, V the velocity, t the time and S0 the initial distance (initial displacement).
From this you can calculate t, if that's what you want.
The magnetic field 0. 02 m from a wire is 0. 1 T. What is the magnitude of the magnetic field 0. 01 m from the same wire? 0. 01 T 0. 05 T 0. 1 T 0. 2 T.
The magnitude of the magnetic field 0.01 m from the same wire is 0.2 T.
Given to us:
Magnetic field, [tex]B_1 = 0.1\ T[/tex]
Radius of wire, [tex]R_1 = 0.02\ m[/tex]
To find out the magnitude of the magnetic field 0. 01 m from the same wire, we need to find out current first. we will use the formula,
[tex]B = \dfrac{\mu_oI }{2\pi R},\\\rn\\where,\\B= magnetic\ field\\\mu_o = 4\pi\times 10^{-7} m\cdot kg\cdot s^{-2} A^{-2}\ is\ the\ magnetic\ constant\\I= current\\R= radius\ of\ the\ wire[/tex]
Putting the values,
[tex]B_1 = \dfrac{\mu_oI }{2\pi R_1},\\\rn\\\\0.1= \dfrac{4\times \pi \times 10^{-7}\times I}{2\times \pi\times0.02}\\\\I=10,000\ A[/tex]
Now, for [tex]B_2[/tex]
[tex]B_2 = \dfrac{\mu_oI }{2\pi R_2},\\\rn\\\\B_2= \dfrac{4\times \pi \times 10^{-7}\times 10,000}{2\times \pi\times0.01}\\\\B_2= 0.2\ T[/tex]
Hence, the magnitude of the magnetic field 0. 01 m from the same wire is 0.2 T.
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in what direction will the seesaw rotate and what will the sign of the angular acceleration be?
Answer:
It can rotate in any direction. The sign of the angular acceleration depends on how you set the reference system, it can be both negative or positive.
A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2. 15, 2. 05, 02. 02, 02. 7. Use the table to answer the questions. What is the fastest time trial for the first quarter checkpoint? seconds What is the slowest time trial for the first quarter checkpoint? seconds What is the range of times measured for this checkpoint? seconds.
The fastest time trial for the first quarter checkpoint is 2.02 s.
The slowest time trial for the first quarter checkpoint is 2.7 s.
The range of the times measured for the checkpoint is 0.68 s.
The given parameters;
Time for quarter checkpoint, = 2.15, 2.05, 2.02, 2.7The fastest time trial for the first quarter checkpoint is the least measured time value.
fastest time trial = least time measured
fastest time trial = 2.02 s
The slowest time trial for the first quarter checkpoint is the highest measured time value.
slowest time trial = 2.7 s
The range of the times measured for the checkpoint is difference between the fastest time and slowest time.
Range = fastest time - slowest time
Range = 2.7 - 2.02
Range = 0.68 s
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Answer:
2.02, 2.15, 0.13
Explanation:
A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2.15, 2.05, 2.02, 2.07.
Use the table to answer the questions.
What is the fastest time trial for the first quarter checkpoint?
2.02
seconds
What is the slowest time trial for the first quarter checkpoint?
2.15 seconds
What is the range of times measured for this checkpoint?
0.13 seconds
A certain satellite is in circular orbit about the earth at an altitude of 550km. If the satellite makes a revolution in 110minutes, calculate its orbital speed
Answer:
Approximately [tex]6.59\times 10^{3}\; \rm m\cdot s^{-1}[/tex].
Explanation:
Look up the radius of the earth: approximately [tex]6.371 \times 10^{3}\; \rm km[/tex].
The radius of the orbit of this satellite is the of the radius of the earth (at ground level) plus the height of the satellite relative to ground level:
[tex]\begin{aligned}r &\approx 6.371 \times 10^{3}\; {\rm km} + 550\; {\rm km} \\ &= 6.921 \times 10^{3}\; \rm km\end{aligned}[/tex].
Convert the units of both distance and time to standard units:
Orbital radius:
[tex]\begin{aligned}r &\approx 6.921 \times 10^{3}\; {\rm km} \\ &= 6.921 \times 10^{3}\; {\rm km} \times \frac{10^{3}\; \rm m}{1\; \rm km} \\ &= 6.921 \times 10^{6}\; \rm m\end{aligned}[/tex].
Orbital period:
[tex]\begin{aligned}t &= 110\; \text{minute} \\ &= 110\; \text{minute} \times \frac{60\; \text{second}}{1\; \text{minute}} \\ &= 6.6 \times 10^{3}\; \text{second}\end{aligned}[/tex].
Calculate the circumference of this orbit:
[tex]\begin{aligned}& 2\, \pi\, r \\ \approx\; & 2 \, \pi \times 6.921 \times 10^{6}\; {\rm m} \\ \approx\; & 4.35 \times 10^{7}\; \rm m\end{aligned}[/tex].
Calculate the orbital speed (tangential) of this satellite:
[tex]\begin{aligned}v &= \frac{2\, \pi\, r}{t} \\ &\approx \frac{4.35 \times 10^{7}\; \rm m}{6.6 \times 10^{3}\; \rm s} \\ &\approx 6.6 \times 10^{3}\; \rm m \cdot s^{-1}\end{aligned}[/tex].
the purpose of many scientific investigations is to test a {n}
Answer:
Scientific investigation is a quest to find the answer to a question using the scientific method.
Explanation:
the scientific method is a systematic process that involves using measurable observations to formulate, test or modify a hypothesis.
A ball falling from a height of 5 m was caught at some height after being reflected off the floor. Find the magnitude of the movement of the ball. If the distance traveled is equal to 7 meters.
Answer:
3m
Explanation:
I can't make a drawing right know, but it's all around vectors. First of all the ball travels down 5 meters, so we have a vector pointing down, then it is reflected and travels 2 meters up. You have to sum the vectors so you obtain that the distance traveled is: 5-2 meters = 3m
The app claimed it has "Forbidden text"
What is the final temperature if it requires 5000 J of heat to warm 2.38892 x10-2 kg of water that starts at 5oC? Remember Cp for water is 4186 J/kgC
The final temperature of water is equal to 50.9999°C
Given the following data:
Mass = [tex]2.38892 \times 10^{-2}\;kg[/tex]Quantity of heat = 5000 J Specific heat capacity of water = 4186 J/kg°CTo determine the final temperature of water:
Mathematically, quantity of heat is given by the formula;
[tex]Q=mc\theta[/tex]
Where:
Q represents the quantity of heat.m represents the mass of an object.c represents the specific heat capacity.∅ represents the change in temperature.Substituting the given parameters into the formula, we have;
[tex]5000=2.38892 \times 10^{-2}\times 4186 \times \theta\\\\5000=100.0001912 \theta\\\\ \theta=\frac{5000}{100.0001912} \\\\ \theta=49.9999^{\circ}C[/tex]
For the final temperature:
[tex]\theta = T_2 - T_1\\\\T_2 = \theta+T_1\\\\T_2 = 49.9999 + 50[/tex]
Final temperature = 50.9999°C
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Mars is 7.83x10^10m [^10 is an exponent] from planet earth. The planet Earth is 5.98x10^24kg [^24 is an exponent] while Mars has a mass of 6.42x10^23kg [^23 is an exponent]. What is the gravitational attraction between the two planets? G=6.67×10^-11 (-11 is an exponent)
Answer:
Approximately [tex]4.18 \times 10^{16}\; \rm N[/tex].
Explanation:
Consider two objects of mass [tex]m_{1}[/tex] and [tex]m_{2}[/tex]. Let [tex]r[/tex] denote the distance between the center of mass of each object. Let [tex]G[/tex] denote the gravitational constant. ([tex]G \approx 6.67 \times 10^{-11}\; {\rm m^{3}\cdot kg^{-1}\cdot s^{-2}}[/tex].)
By Newton's Law of Universal Gravitation, the size of gravitational attraction between these two objects would be:
[tex]\begin{aligned}F &= \frac{G\, m_{1}\, m_{2}}{r^{2}}\end{aligned}[/tex].
In this question, [tex]m_{1} = 5.98\times 10^{24}\; {\rm kg}[/tex] and [tex]m_{2} = 6.24 \times 10^{23}\; {\rm kg}[/tex] are the mass of the two planets. The distance between the two planets is [tex]r = 7.83 \times 10^{10}\; \rm m[/tex] (approximately the same as the distance between the center of mass of planet Earth and the center of mass of Mars.)
Apply Newton's Law of Universal Gravitation to find the size of gravitational attraction between the two planets:
[tex]\begin{aligned}F &= \frac{G\, m_{1}\, m_{2}}{r^{2}} \\ &= \frac{1}{(7.83 \times 10^{10}\; {\rm m})^{2}} \\ &\quad \times (6.67 \times 10^{11}\; {\rm m^{3} \cdot kg^{-1} \cdot s^{-2}}) \\ &\quad \times (5.98 \times 10^{24}\; {\rm kg}) \\ &\quad \times (6.42 \times 10^{23}\; {\rm kg}) \\ &\approx 4.18 \times 10^{16}\; {\rm kg \cdot m \cdot s^{-2}} \end{aligned}[/tex].
Since [tex]1\; {\rm kg \cdot m \cdot s^{-2}} = 1\; {\rm N}[/tex], the size of gravitational attraction between the two planets would be approximately [tex]4.18 \times 10^{16}\; {\rm N}[/tex].
how much force would be needed to push a box weighing 30 N up a ramp that ahas an ideal mechanical advantage of 3
Answer:
60n would be the needed force
The force needed to push a box weighing 30 N up a ramp that has an ideal mechanical advantage of 3 is equal to 10 N.
What is the mechanical advantage?The mechanical advantage can be described as the ratio of the input force to the output force. The mechanical advantage of any machine can be determined by the ratio of the forces involved to do the work.
The ratio of the resistance force to the effort is called the actual mechanical advantage which will be comparatively less. The efficiency of a machine is always determined by equating the ratio of its output to its input.
The efficiency of the machine is equal to the ratio of the actual mechanical advantage (M.A.) and theoretical mechanical advantage. Mechanical advantage can be defined as the force produced by a machine to the force applied to it.
Given the load = 30 N and the ideal mechanical advantage = 3
Mechanical advantage = Load/ Effort
Input force or effort = Load/ M.A.
Force = 30/3
Input Force = 10 N
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The normal force of a parked car on a level surface is 15,000 Newtons. What is the force of the car?
Answer:
The force of the car is 15000N.
Explanation:
The unit of force is Newtons (N), so based on the question, the force is 15000 Newtons.
please help quickly,the picture and choices are above.
Answer:
C because it make sens
C the light wave traveled through ice and then through a Dimond.
4 What type of circuit is described by each of the following statements?
Answer series or parallel
a All components are connected end-to-end.
b. The current in the circuit divides so that some flows through one component
and the rest through another component
Two lamps are connected side by side so that each lights brightly
d The current has the same valur everywhere in the circuit
C
What are the three different social perspectives on sport
Answer:
functionalist theory,feminist theory. discipline of sociology
[tex] \huge{ \mathrm{question \hookleftarrow}}[/tex]
Calculate equivalent resistance of two resistors R1 and R2 in parallel where,
[tex] \sf R_ 1 = (6 \pm0.2 )\: \: ohms[/tex]
[tex] \sf R_ 2 = (3 \pm0.1 )\: \: ohms[/tex]
[tex]it \: was \: helpful \: to \: you[/tex]
# be careful#
The gravitational force between two asteroids is 6.2 × 108 N. Asteroid Y has three times the mass of asteroid Z.
If the distance between the asteroids is 2100 kilometers, what is the mass of asteroid Y?
Answer:
1.1x10¹⁶ kg
Explanation:
Let m = mass of asteroid y.
Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.
Given:
F = 6.2x10⁸ N
d = 2100 km = 2.1x10⁶ m
Note that
G = 6.67408x10⁻¹¹ m³/(kg-s²)
The gravitational force between the asteroids is
F = (G*m*(m/3))/d² = (Gm²)/(3d²)
or
m² = (3Fd²)/G
= [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))
= 1.229x10³² kg²
m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)
29) A cheetah can accelerate from rest to 25
m/s in 6 s. Assuming that the cheetah moves
with constant acceleration, what distance does
it cover in the first 3 s
Answer:
[tex]\huge\boxed{\sf S = 18.75 \ meters}[/tex]
Explanation:
Given Data:
Initial Velocity = Vi = 0 m/s (rest)
Final Velocity for 6 seconds = Vf = 25 m/s
Time (1) = T1 = 6 seconds
Time (2) = T2 = 3 seconds
Required:
Distance for 3 seconds = S = ?
Solution:
For 6 seconds, the acceleration will be:
[tex]\displaystyle a = \frac{Vf-Vi}{t} \\\\a = \frac{25 - 0}{6} \\\\a = 25 / 6\\\\\boxed{a = 4.167 \ m/s^2}[/tex]
Since, acceleration is constant, it will be the same at 3 seconds as well.
Using second equation of motion to find Distance (S) with time being 3 seconds:
[tex]\displaystyle S= Vit+\frac{1}{2} at^2\\\\S = (0)(3)+ \frac{1}{2} (4.167)(3)^2\\\\S = \frac{1}{2} (4.167)(9)\\\\S = \frac{37.5}{2} \\\\\boxed{S = 18.75 \ meters}\\\\\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807A 2.0 kg block rests on a level surface. The coefficient of static friction is, and the coefficient of kinetic friction is A horizontal force, X, is applied to the block. As X is increased, the block begins moving. Describe how the force of friction varies as X increases from the moment the block is at rest to when it begins moving. Indicate how you could determine the force of friction at each value of X―before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
ps. I had to change F to X because of brainly.
By Newton's second law, the net force acting on the block in the vertical direction is
∑ F [ver] = n - mg = 0
where n = magnitude of normal force and mg = weight of the block. It follows that n = mg.
When the block is at rest, the applied force X will not be enough to move the box until it can overcome the maximum mag. of static friction. If µ[s] is the coefficient of static friction, then the maximum mag. of the frictional force is
f = µ[s] n = µ[s] mg
The net horizontal force would be
∑ F [hor] = X - µ[s] mg = 0
so a minimum force of X = µ[s] mg is required to get the block moving. Any mag. smaller than this and the block stays at rest/in equilibrium.
Once the mag. of X exceeds µ[s] mg, the block will begin to move. At that point, if the coefficient of kinetic friction is µ[k], then the net force on the block is
∑ F [hor] = X - µ[k] mg = 0
so a minimum force of X = µ[k] mg would be needed to keep the block moving at constant speed, or otherwise X = µ[k] mg + ma if the block is accelerating with mag. a.
The principles here are captured in the attached plot.