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emmasim [6.3K]
3 years ago
14

Part complete during a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s

away from the wall. if the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball?
Physics
1 answer:
Inga [223]3 years ago
8 0

The average force applied to the ball= 106.7 N

Explanation:

Force is given by

f= ΔP/t

ΔP= change in momentum= m Vf- m Vi

m= mass =0.2 kg

Vf= final velocity= 12 m/s

Vi=initial velocity= -20 m/s ( negative because it is going towards the wall which is treated as negative axis)

t= time= 60 ms= 0.06 s

now ΔP= 0.2 [ 12-(-20)]

ΔP=0.2 (32)=6.4 kg m/s

now force F= ΔP/t

F= 6.4/0.06

F=106.7 N

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A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction i
IrinaVladis [17]
A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
W=Fd = (46.3 N)(4.25 m)=196.8 J

b) The work done by the frictional force against the motion of the block is equal to:
W_f =  -F_f d =- (\mu mg) d =-(0.609)(4.14 kg)(9.81 m/s^2)(4.25 m)=
=-105.1 J
Part of these 105.1 Joules of work becomes increase of thermal energy of the block (\Delta E_B), and part of it becomes increase of thermal energy of the floor (\Delta E_F). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
\Delta E_F = 105.1 J - \Delta E_B = 105.1 J-38.2 J=66.9 J

c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
W_{net}=W-W_f=196.8 J-105.1 J=91.7 J
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
W_{net} = \Delta K
So, we have \Delta K=+91.7 J


5 0
3 years ago
Gas pressure in a closed system is caused by___. A. Heat given off by constant collisions.. B. Compression of the gas. C. Result
jekas [21]
Gas pressure in a closed system is caused by resultant force of all combines collisions. The correct option among all the options that are given in the question is the third option or option "C". The other options can easily be ignored. I hope the answer comes to your help.
8 0
3 years ago
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A dragster car does the ¼ mile in 5.32 seconds. (1 mile = 5280 feet)
svet-max [94.6K]

Answer:

well most top fuel dragsters go from 0-60 in under a second, and can go 0-100 in like 1.5 seconds so the top speed would be about 290 mph- 300

Explanation:

the fastest top fuel dragster went to 336mph for 1/4mile

4 0
3 years ago
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Pun zapato de golf tiene 10 tacos cada uno con un área de 0.01 pulgadas en contacto con el piso suponga que caminar hay un insta
Mars2501 [29]

Answer:

Las presión ejercida por los tacos sobre el suelo es 900 libras por pulgada cuadrada.

Explanation:

Según la Física, comprendemos que la presión es igual a la fuerza dividida por área. En este caso, el peso de la persona dividida por área total de los tacos de los zapatos de golf suponiendo una distribución uniforme de la fuerza. Es decir:

\sigma = \frac{W}{n\cdot A_{T}} (Eq. 1)

Donde:

\sigma - Presión, medida en libras por pulgada cuadrada.

W - Peso de la persona, medida en libras.

n - Cantidad de tacos en los zapatos, adimensional.

A_{T} - Área del taco, medida en pulgadas cuadradas.

Si conocemos que W = 180\,lb, n = 20 y A_{T} = 0.01\,in^{2}, la presión es:

\sigma = \frac{180\,lb}{(20)\cdot (0.01\,in^{2})}

\sigma = 900\,psi

Las presión ejercida por los tacos sobre el suelo es 900 libras por pulgada cuadrada.

3 0
3 years ago
In Part I, the independent variable, the one that is intentionally manipulated, is blank.
Feliz [49]

Answer: part 1 is mass and part 2 is density

Explanation: I just did it and got it right

6 0
3 years ago
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