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3 years ago

15

How would I answer this equation? E= mc^2 for m

1 answer:

zubka84 [21]3 years ago

5 0

The E stands for energy. M stands for mass c squared stands for speed of light squared

energy= mass speed of light squared

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A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5

harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

$E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\$

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

$P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\$

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0

3 years ago

A result of chemical change is

Tcecarenko [31]

We're u can never put it back together

4 0

3 years ago

if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball

grigory [225]

It would exert the same back right?

3 0

3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa

maw [93]

Answer:

Explanation:

Given

Initial speed $u=60\ mi/hr\approx 88\ ft/s$

distance traveled before coming to rest $d_1=120\ ft$

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$0-(88)^2=2\times a\times 120---1$

for $u_2=80\ mi/hr\approx 117.33\ ft/s$

using same relation we get

$0-(117.33)^2=2\times a\times (d_2)----2$

divide 1 and 2 we get

$(\frac{88}{117.33})^2=\frac{120}{d_2}$

$d_2=213.32\ ft$

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0

3 years ago