1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alexandr1967 [171]
3 years ago
12

A marble and a basketball are rolling toward you at the same velocity. If the same amount of force is used to stop both objects,

what is true about the amount of time needed to stop those objects?
Physics
1 answer:
Ira Lisetskai [31]3 years ago
6 0

Answer:

The amount of time needed to stop the marble is much less than the amount of time needed to stop the basketball.

Explanation:

To explain this question we can use the equation:

Change in momentum = Force x Time

Where momentum = mass x Velocity

In the case of the basketball, it has a much larger mass than the marble, meaning it has much more momentum than the marble. This means that with the same amount of force, it will need much more time to stop.

Likewise, the marble has less mass than the basketball meaning it will have less momentum. This means that with the same amount of force, it will require less time to stop.

Logically this also makes sense if we think about it in a real life scenario. If we try to stop a marble with our finger, we can stop it much faster then if we try to stop a basketball with our finger because of the difference in mass between the two objects, therefore the marble will require much less time to be stopped than the basketball.

Hope this helped!

You might be interested in
A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s
strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v

Where:

m_{M}, m_{S} - Masses of the small meteorite and the communication satellite, measured in kilograms.

v_{M}, v_{S} - Speeds of the small meteorite and the communication satellite, measured in meters per second.

v - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}

If m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s} and v_{S} = 0\,\frac{m}{s}, the final speed is now calculated:

v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}

v = 0.1\,\frac{m}{s}

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

K_{S} + K_{M} - K - Q_{disp} = 0

Q_{disp} = K_{S}+K_{M}-K

Where:

K_{S}, K_{M} - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

K - Kinetic energy of the satellite-meteorite system, measured in joules.

Q_{disp} - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]

Given that m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s}, v_{S} = 0\,\frac{m}{s} and v = 0.1\,\frac{m}{s}, the dissipated heat is:

Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]Q_{disp} = 200000\,J

Q_{disp} = 200\,kJ

The energy coverted to heat is 200 kilojoules.

4 0
3 years ago
Suppose an individual is lying on his stomach with sheets of paper stacked on his back. If each sheet of paper has a mass of 0.0
AURORKA [14]

Answer:

N = 177843 sheets

Explanation:

We are given;

Mass;m = 0.0035 kg

Pressure; p = 101325 pa = 101325 N/m²

L = 0.279m

W = 0.216m

The weight of N sheets is N(mg)

Where;

m is the mass of one sheet

N is number of sheets

g is the acceleration due to gravity.

The pressure equals weight divided by the area on which the weight presses:

Thus,

p= F/A = Nmg/(L•W)

Therefore, making N the subject;

N = pLW/(mg)

N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)

N = 177843

5 0
3 years ago
Describe 2 ways you could increase the efficiency of a household central heating system
chubhunter [2.5K]

Answer:

Arrange an annual service. Treat your boiler like your car. ...

Keep your boiler clean. ...

Bleed your radiators. ...

Top up the pressure. ...

Use a Powerflush. ...

Insulate your pipes. ...

Turn the heating on. ...

If all else fails…

Explanation:

6 0
3 years ago
1.A student attaches a string to a puck on a frictionless air table, and pulls with a constant force on the puck. a. Draw the fo
Mrrafil [7]

a)

for the puck :

F = force applied in the direction of pull

N = normal force on the puck in upward direction by the surface of table

W = weight of the puck in down direction due to force of gravity


b)

along the vertical direction , normal force balance the weight of the puck , hence the net force is same as the force of pull F .

so  F = ma                                    where m = mass of puck  , a = acceleration

Fnet = F


c)

since the net force acts in the direction of force of pull F , hence the puck accelerates in the same direction .

6 0
2 years ago
A 22.0-kg child is riding a playground merrygo-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he i
Tems11 [23]

Answer:

F=480.491 N

Explanation:

Given that

mass ,m = 22 kg

Angular speed ω = 40 rev/min

\omega=\dfrac{2\pi \times 40}{60}\ rad/s

ω =4.18 rad/s

The radius  r= 1.25 m

We know that centripetal force is given as

F=m ω² r

Now by putting the values in the above equation we get

F=22\times 4.18^2\times 1.25\ N

F=480.491 N

Therefore the centripetal force on the child will be 480.491 N.

6 0
3 years ago
Other questions:
  • Find the length of an arc with a radius of 6.0m swept across 2.5 radians.
    9·1 answer
  • 10 Points!
    7·2 answers
  • When white light is viewed through sodium vapor in a spectroscope the spectrum is continuous?
    7·2 answers
  • A force that comes from the action of earth's Gravity is called
    6·2 answers
  • A mass slider m = 0.200 kg rests on a frictionless horizontal air rail connected to a spring with a force constant k = 5.00 N /
    7·1 answer
  • Part A Which of the following statements is/are true? Check all that apply.
    12·1 answer
  • What can be increased in order to increase the magnetic field?
    12·2 answers
  • The melting of a glacier is an example of the interactions among which of Earth’s spheres?
    15·1 answer
  • The low point of a transverse wave is called a
    14·1 answer
  • Two identical light bulbs, each with resistance R = 2  are connected to a source with E = 8 V and negligible internal resistanc
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!