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2 years ago

A man pushes a 10kg box with a constant acceleration of 5m/s2. What force is applied to the box?

2 answers:

8 0

FORMULA

$\boxed {F = m \times a}$

F = force
m = mass
a = acceleration

Using the formula

$F = 10 \times 5$

Multiply

$\boxed {\textsf {F = 10 N}}$

Diano4ka-milaya [45]2 years ago

8 0

Depends on how long you push it.

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What type of friction prevents a pile of rocks from falling apart?

BlackZzzverrR [31]

A because it some type of friction

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2 years ago

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

2 years ago

When bouncing a ball, the bouncing motion results in the ball ____________.

Alekssandra [29.7K]

Answer: "B" Changing Position

Great Question!

Explanation: When a ball bounces to the ground it hits the ground with some energy. The amount of energy with which it hits the ground is kinetic energy. When it comes in the contact with the ground kinetic energy gets converted into potential energy. This potential energy again gets converted into kinetic energy and balls moves again from the ground and bounces multiple times. So, the ball ends up changing position

8 0

2 years ago

Can someone help me please

Evgen [1.6K]

The last one: meter

5 0

2 years ago

Read 2 more answers

Variables in physics often include a subscript. What are subscripts used for in physics?

Viefleur [7K]

Answer:

C.) To indicate different versions of the same variable.

Explanation:

Variables in physics often include a subscript. These subscripts are used for indicating different versions of the same variable in physics.

Basically, subscripts are used to represent the beginning (initial) and ending (final) position or point of a variable in physics.

For example, we would look at Gay Lussac' Law of gases.

Gay Lussac law states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Gay Lussac's law is given by;

$PT = K$