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3 years ago

How do i calculate this?

Physics

1 answer:

Lesechka [4]3 years ago

8 0

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

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.A cart rolling down an incline for 5.0 seconds has an acceleration of 1.6 m/s2. If the cart has a beginning speed of 2.0 m/s, a

Ilia_Sergeevich [38]

Use the formula,

$\Delta x=v_it+\dfrac12at^2$

where $\Delta x$ is the cart's displacement (from the origin), $v_i$ is its initial speed, $a$ is its acceleration, and $t$ is time.

$\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2$

$\implies\boxed{\Delta x=30\,\mathrm m}$

Alternatively, since acceleration is constant, we have

$\dfrac{v_f+v_i}2=\dfrac{\Delta x}t$

That is, we have these two equivalent expressions for average velocity, where $v_f$ is the cart's final velocity. Solve for $\Delta x$ :

$\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}$

$\implies\boxed{\Delta x=30\,\mathrm m}$

8 0

3 years ago

A gas within a piston–cylinder assembly undergoes a

IrinaVladis [17]

Answer:

$W_{net}= -280.5 kJ$

$Q_{23}= 80.0 kJ$

Explanation:

The net work of the cycle is the sum of works in each process.

For the first process 1-2: Let's apply the work definition.

$W=\int pdV$ (1)

Now, we need the pressure. We know that pV=C, where C is a constant. Then $p_{1}V_{1}=10^{5}2=2*10^{5} [J]$

So $p=\frac{2*10^{5}}{V}$

Let's put p in (1):

$W_{12}=\int \frac{2*10^{5}}{V}dV=2*10^{5}\int \frac{1}{V}dV$

$W_{12}=2*10^{5}(ln(V_{2})-ln(V_{1}))=2*10^{5}(ln(0.2)-ln(2))=-460.5 kJ$

Using the first law of thermodynamics we can find Q.

$Q_{12}-W=\Delta U$

$Q_{12}=W+\Delta U=-460.5 + 100= -360.5 kJ$

Second process 2-3

In this case, we have a constant volume, so the work done here is 0.

$W_{23}=0$

Third process 3-1

$W_{31}=\int pdV=p\int dV=p(V_{2}-V_{1})$

$W_{31}=10^{5}(2-0.2)=180 kJ$

Finally, the net work is:

$W_{net}=W_{12}+W_{23}+W_{31}= -280.5 kJ$

By the conservation of energy:

$(Q_{12}+Q_{23}+Q_{31})-(W_{net})=\Delta E=0$

Because there is no change in total energy.

So:

$Q_{23}=W_{net}-Q_{12}= 80.0 kJ$

It is a refrigerator because the net work is negative, it means it consumes energy.

I hope it helps you!

3 0

3 years ago

An object that is in free fall seems to be ______

Tomtit [17]

The answer would be weightless. The occurrence of "weightlessness" happens when there is no force of support on your body. When your body is efficiently in "free fall", quickening downward at the acceleration of gravity, then you are not being maintained.

3 0

2 years ago

Answer with step by step

Archy [21]

If 'C' and 'R' represent capacilance and resistance respectively then the dimensions of RC are:

A:M^0L^0T^2
B:M^0L^0T^1
C:Ml^1
D:Ml^-1T^2
E;None of these

7 0

3 years ago

(This is for other people with this Question i hope you find this when you need help) Need answer for this Help!!!

Radda [10]

Answer:

0.48N

Explanation:

according to the second law of motion

force=mass×acceleration

the mass in this question is 0.025,the acceleration 19

therefore f=0.025×19

=0.48N

I hope this helps

7 0

2 years ago