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3 years ago

10

A bullet moving at 500 m/s has a kinetic energy of 2500 J and a potential energy of 0.5 J at a certain moment. What is the bulle

ts mass and how high above the ground is it at the moment

1 answer:

Nataly_w [17]3 years ago

3 0

Half mv squared equaled mgh
.5(500)power 2=9.8h
h=12755m

plug h in to solve for m
*m's cancel out

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A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the

Dima020 [189]

Explanation:

Given that,

Velocity of the proton in lab frame $u_{x} = 0.6c$

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

$u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}$

$u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}$

$u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})$

$u'=0.385c$

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)$

Where, Proton mass energy = m₀c²

Put the value into the formula

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)$

$K.E=234.57\ MeV$

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)$

$K.E=78.366\ MeV$

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

$P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}$

We know that,

Proton mass energy = m₀c²

$m_{0}=\dfrac{938.28}{c^2}$

$P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}$

$P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}$

$P_{obs}=150.69\ MeV/c$

Hence, This is required solution.

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3 years ago

What force would be required to accelerate a 1,100 kg car to 0.5 m/s2?

Klio2033 [76]

Force =mass x acceleration =1100 x0.5 =550

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3 years ago

Calculate the KE of a 3.1 kg toy cart that moves at 4.8 m/s. Calculate the KE of the same cart at twice the speed.

lisov135 [29]

When v= 4.8 m/s:

E.K=mv²/2
=3,1(4.8)²/2
=35.712J

when v= 2(4.8):

E.K=mv²/2
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4 years ago

22. (I) A car slows down from 28 m/s to rest in a distance of

lidiya [134]

Answer:

a = -8.912 m/s²

Explanation:

Given,

The initial velocity of the car, u = 28 m/s

The final velocity of the car, v = 0

The distance traveled by car, d = 88 m

The velocity displacement relation is given by the formula

v = d/t

∴ t = d/v

Substituting in the above values in the given equation

t = 88/28

= 3.142 s

The acceleration is given by the formula

a = (v-u)/t

= (0 - 28)/3.142

= -8.912 m/s²

The negative sign is that the car is decelerating.

Hence, acceleration a = -8.912 m/s²

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4 years ago

an open rectangular tank 1.0 meter wide and 2.0 meter long contains gasoline to a depth of 1.5 meters. if the height of the tank

Alekssandra [29.7K]

The maximum horizontal acceleration is solved to be 4.91 m/s^2

<u />

<h3>What is acceleration?</h3>

This is a term in physics that refers to change in velocity with time. Acceleration can be positive or negative or zero,

<u />

<u>Given data</u>

open rectangular tank

width = 1.0 m

length = 2.0 m

gasoline depth level = 1.5 m

height of the tank = 2.0 m

<h3>How to solve for maximum horizontal acceleration</h3>

lets represent the volume with x, y, and z such that z corresponds to height, y corresponds to wide or width, and x corresponds to length.

dz / dy = - ay / ( g + az )

where

ay = acceleration in the y direction

az = acceleration in the z direction which is equal to zero

g = acceleration due to gravity

plugging in az = 0

dz / dy = - ay / ( g )

ay = - ( dz / dy ) * g

to prevent spilling

dz / dy = - ( 1.5 - 1 ) / 1

dz / dy = - 0.5

ay = - ( - 0.5 ) * 9.81

ay = 4.91 meters per second square

Hence the maximum horizontal acceleration is solved to be 4.91 m/s^2

Read more on acceleration here: brainly.com/question/605631

#SPJ1

7 0

2 years ago