A rock falls for 3 seconds. What was its velocity when it hit the ground?

A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2

SVEN [57.7K]

Answer:

$K=512J$

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass $m_1$ has an initial velocity $v_{1i}$ and a final velocity $v_{1f}$ and the block of mass $m_2$ has an initial velocity $v_{2i}$ and a final velocity $v_{2f}$ then the initial and final momentum of the system will be:

$p_i=m_1v_{1i}+m_2v_{2i}$

$p_f=m_1v_{1f}+m_2v_{2f}$

Since momentum is conserved, $p_i=p_f$ , which means:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We know that the block is brought to rest by the collision, which means $v_{2f}=0m/s$ and leaves us with:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}$

which is the same as:

$v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}$

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

$v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s$

So kinetic energy of the bullet as it emerges from the block will be:

$K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J$

6 0

3 years ago

Suppose that the design parameters of the satellite's control system require that the angular velocity of the satellite not exce

Stells [14]

Answer:

v=0.04m/s

Explanation:

To solve this problem we have to take into account the expression

$\omega=\frac{v}{r}$

where v and r are the magnitudes of the velocity and position vectors.

By calculating the magnitude of r and replacing w=0.02rad/s in the formula we have that

$|r|=\sqrt{(-1.18m)^2+(-0.9m)^2}=2.01m\\\\v=\omega r=(0.02\frac{rad}{s})(2.01m)=0.04\frac{m}{s}$

the maximum relative velocity is 0.04m/s

hope this helps!!

6 0

3 years ago

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Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe

Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

$T^{2} \alpha R^{3}$

⇒ $T^{2} = KR^{3}$ where is k is the proportionality constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence $T_{1} =365.5 days$

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence $R_{1} =1 AU$

The distance of venus from sun is 0.723 AU i.e $R_{2} =0.723$

From keplers law we know that- $\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }$

⇒ $T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }$

Putting the values mentioned above we get-

$T_{2} ^{2} =50,350.132851075$

⇒ $T_{2} =\sqrt{50,350.132851075}$

⇒ $T_{2} = 224.388352752710 days.$

Hence the time period of venus is 224.388352752710 days

7 0

4 years ago

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Please help with this problem.

pochemuha

Answer:

its Y because the dot that represents it is close to S

6 0

4 years ago

Nutritionists use the ______ to measure how much energy we get from foods.

Natali [406]

<span>It is measured in calories </span>

6 0

3 years ago

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