Kinetic energy portfolio in part 2 the independent changes to----?

A car is moving from rest and the velocity increases to 30 m/s in 4 seconds. Calculate its acceleration.

vaieri [72.5K]

Answer:

7.5 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 30 m/s

t = 4 s

Find: a

v = at + v₀

(30 m/s) = a (4 s) + (0 m/s)

a = 7.5 m/s²

8 0

3 years ago

What is the equivalent of 0° C in Kelvin? Help will give brainiest

givi [52]

Answer:

273.15 dat is the answer im pretty sure

8 0

3 years ago

A car with four passengers will have a shorter braking distance than a car with one braking distance. True or false. Why?

Valentin [98]

Answer:

The given statement is a True statement

Explanation:

All cars and trucks have load capacities marked on the jamb of the driver's door, as well as the owner's manual. This is very significant for braking distance.

A heavier object will require much more braking force and distance to stop, due to New ton's law of motion. “ An object in motion will tend to stay in motion “ .A heavy object will remain in motion longer and require much more force to stop. Four wheel disc brakes are great compared to front disc and rear drum configuration, in dissipating heat, and stopping more efficiently.

6 0

3 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

3 years ago

F all of the energy in a falling object's gravitational potential energy store is transferred to its kinetic energy store by the

stepladder [879]

Answer:

The options are not shown, so let's derive the relationship.

For an object that is at a height H above the ground, and is not moving, the potential energy will be:

U = m*g*H

where m is the mass of the object, and g is the gravitational acceleration.

Now, the kinetic energy of an object can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.

Uinitial = Kfinal.

m*g*H = (1/2)*m*v^2

v^2 = 2*g*H

v = √(2*g*H)

So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.

5 0

2 years ago