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harkovskaia [24]
3 years ago
5

Jerome plays middle linebacker for souths varsity football team. In a game against cross-town rival north, he delivered a hit to

north's 82-kg running back, changing his eastward velocity of 5.6 m/s into a westward velocity of 2.5 m/s. Determine that momentum change of the running back
Physics
1 answer:
Makovka662 [10]3 years ago
7 0

The momentum change of the running back is - 664.2 kg m/s or 664 west.

<u>Explanation:</u>

Momentum is defined as the change in velocity of any object along with its mass. So mathematically, momentum can be derived using the product of mass with the change in velocity.

           \text { Momentum change }=m \times \Delta v

As here mass is given as 82 kg and the initial velocity was 5.6 m/s and final velocity is 2.5 m/s.

      Initial Momentum = m \times \text { initial velocity }=82 \times 5.6=459.2 \mathrm{kg} \mathrm{m} / \mathrm{s}

      Final Momentum = m \times \text { final velocity }=82 \times(-2.5)=-205 \mathrm{kg} \mathrm{m} / \mathrm{s}

      Momentum change = Final Momentum - Initial Momentum

      Momentum change = - 205 - 459.2 = - 664.2 kg m/s or 664 west

Thus, the changing momentum is -664.2 kg m/s. The negative sign indicates that the momentum is acting in the opposite direction on changing in the direction of velocity.

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It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
TEA [102]

Answer:

v=346.05\ m.s^{-1}

Explanation:

Given:

initial temperature of the lead bullet, T_i=43^{\circ}C

latent heat of fusion of lead, L_f=2.32\times 10^4\ J.kg^{-1}

melting point of lead, T_m=327.3^{\circ}C

We have:

specific heat capacity of lead, c=129\ J.kg^{-1}.K^{-1}

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)

\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f

\frac{1}{2} m.v^2=m(c.\Delta T+L_f)

\frac{v^2}{2} =129\times(327.3-43)+23200

v=346.05\ m.s^{-1}

3 0
3 years ago
A bus travels a distance of 120 km with a speed of 40km per hour and returns with a speed of 30km per hour calculate the average
Vsevolod [243]

Answer:

35 km/hr

Explanation:

Average speed = (total of the speed)/(the sets of speeds given)

Direction does not matter in this instance since speed is only magnitude,

Average speed = (30 + 40)/2

Average speed = 70 ÷ 2

= 35 km/hr

3 0
3 years ago
A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea
viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}                           (1)

time taken = 5 years

We know that :

time = \frac{distance}{speed}

5 = \frac{L''}{v}                                                      (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

Solving the above eqn:

t'' = 20.56 years

4 0
3 years ago
Which is an example of an unstructured activity that promotes resistance training?
aleksandrvk [35]
One possible unstructured activity that promotes resistance training would be climbing playground equimpent - A. 

This is by nature a unstructured ctivity. Furthermore, it promotes resistance training because you're forced to move and pull and push yourself. 
3 0
3 years ago
What stores energy for a quick release in a cell?
luda_lava [24]
It is the mitochondria of a cell that stores energy for a quick release. <span>Mitochondria break down glucose to release the energy for cells to use. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
5 0
3 years ago
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