Sequence eye light enters the eye

Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet of

Harrizon [31]

Answer:

a) 6.57 m/s

b) 53.75 J

c) 6.37 m/s

d) -98.297 J

Explanation:

mass of player = $m_{p}$ = 117.5 kg

speed of player = $v_{p}$ = 6.5 m/s

mass of ball = $m_{b}$ = 0.43 kg

velocity of ball = $v_{b}$ = 26.5 m/s

Recall that momentum of a body = mass x velocity = mv

initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s

initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s

initial kinetic energy of the player = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.5^{2}$ = 2482.187 J

a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.

for this first case that they travel in the same direction, their momenta carry the same sign

$m_{p}$ $v_{p}$ + $m_{b}$ $v_{b}$ = ( $m_{p}$ + $m_{b}$ )v

where v is the final velocity of the player.

inserting calculated momenta of ball and player from above, we have

763.75 + 11.395 = (117.5 + 0.43)v

775.145 = 117.93v

v = 775.145/117.93 = 6.57 m/s

b) the player's new kinetic energy = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.57^{2}$ = 2535.94 J

change in kinetic energy = 2535.94 - 2482.187 = 53.75 J gained

c) if they travel in opposite direction, equation becomes

$m_{p}$ $v_{p}$ - $m_{b}$ $v_{b}$ = ( $m_{p}$ + $m_{b}$ )v

763.75 - 11.395 = (117.5 + 0.43)v

752.355 = 117.93v

v = 752.355/117.93 = 6.37 m/s

d) the player's new kinetic energy = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.37^{2}$ = 2383.89 J

change in kinetic energy = 2383.89 - 2482.187 = -98.297 J

that is 98.297 J lost

3 0

4 years ago

A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla

Yuri [45]

Answer:

$= - 5.65\times 10^{-2} J$

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18 *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

$U =- G \frac{m_1 m_2}{r}$

mass per unit length is given as

$\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m$

calculating potential energy

$dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}$

$=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}$

$=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}$

$=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}$

$=-G*m_1*\sigma ln(1.125)$

$=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)$

$= - 5.65\times 10^{-2} J$

5 0

3 years ago

PLEASE HELP!!!!!!!! NEED DONE SOON!!!

zhannawk [14.2K]

Answer:

the object is decelerating

4 0

3 years ago

Read 2 more answers

How is potential energy unique?

ahrayia [7]

Answer:

is uniquely defined.

Explanation:

The gravitational potential energy gained by an object being lifted is equal to the work done on the object by an applied force, which exactly cancels gravity. The applied force does positive work. When the object is being lifted, the gravitational force does negative work.

3 0

4 years ago

Why are hot springs and geysers is found in areas of present or a past activity

Lady_Fox [76]

Hot springs and geysers are often found in areas of present or past volcanic activity. The hot spring forms when water deep underground is heated by a nearby body of magma or hot rock. The hot water rises and collects in a natural pool; when rising hot water and steam become trapped in a narrow crack, pressure builds until the mixture sprays above the surface as a geyser.

7 0

3 years ago