The first and second laws apply to ________________

How do you state a hypothesis! Branliest

VLD [36.1K]

Answer:

State your hypothesis as concisely, and to the point, as possible. A hypothesis is usually written in a form where it proposes that, if something is done, then something else will occur. Usually, you don't want to state a hypothesis as a question. You believe in something, and you're seeking to prove it.

Explanation:

4 0

3 years ago

Read 2 more answers

How does the kinetic energy from the forward motion of a car traveling at 16 m/s

elena55 [62]

The kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

<h3>What is Kinetic Energy?</h3>

Kinetic energy is simply a form of energy a particle or object possesses due to its motion.

It is expressed as;

K = (1/2)mv²

Where m is mass of the object and v is its velocity.

Given that;

For the first case, velocity v = 16m/s
For the second case, velocity = 8m/s
Let the mass of the car be m

For the first case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (16m/s)²

K = (1/2) × m × 256m²/s²

K = mass × 128m²/s²

For the second case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (8m/s)²

K = (1/2) × m × 64m²/s²

K = mass × 32m²/s²

Comparing the kinetic energy of the car with the same mass but different velocity, we can see that the kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

Learn more about kinetic energy here: brainly.com/question/12669551

#SPJ1

7 0

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

3 years ago

What formula relstes work n power

777dan777 [17]

<span>Raj is trying to make a diagram to show what he has learned about nuclear fusion.
</span>

8 0

3 years ago

During a trial run, race car A starts from rest and accelerates uniformly along a straight level track for a particular interval

kondor19780726 [428]

Answer: car B has travelled 4times as far as Car A

d=vi*t+1/2at^2

No initial velocity so equation becomes;

d=1/2at^2 and the acceleration is the same between both only time is different;

Car A d=1/2a(1)^2

Car B d=1/2a(2)^2

Car A d= 1^2=1

Car B d= 2^2=4

Car B d=4*Car A

So car B has travelled 4 times as far as car A

5 0

3 years ago

The first and second laws apply to _________________ objects.