All categories

3 years ago

What does Einstein's famous equation for nuclear energy, E = mc^2, mean?

Physics

1 answer:

Nezavi [6.7K]3 years ago

8 0

Yup the correct answer is A jus finished the quiz :)

hope i helped^_^

You might be interested in

Calculate the acceleration if you push with a 20-N horizontal force against a 2-kg block on a horizontal friction- free air tabl

Nuetrik [128]

10 meters per second. The formula is F = ma, so to solve we substitute 20 for F, 2 for m and then solve for a. 20 divided by 2 - 10, therefore the acceleration is 10 meters per second.

8 0

2 years ago

A 15.4 kg block is dragged over a rough, horizontal surface by a constant force of 182 N acting at an angle of 24◦ above the hor

Dima020 [189]

Answer:

Explanation:

Considering Work done by friction is asked

Given

mass of block $m=15.4\ kg$

force $F=182\ N$

inclination $\theta =24^{\circ}$

block is displaced by $s=57.6\ m$

coefficient of kinetic friction $\mu _k=0.135$

Friction force $F_r=\mu _kN$

Normal reaction $N=mg-F\sin \theta =15.4\times 9.8-182\sin (24)$

$N=76.9\ N$

Friction Force $F_r=0.135\times 76.9=10.38\ N$

Work done by friction force

$W=f_r\cdot s$

$W=10.38\times 57.6=597.92\ N$

3 0

2 years ago

a student weighs 1200N they are standing in an elevator that is moving downwards at a constant speed of

baherus [9]

Answer:

Elevator That Is Moving Downwards At A Constant Speed Of 4.9 M/S. What Is The Magnitude Of The Net Force Acing On The Student?

This problem has been solved!

This problem has been solved!See the answer

This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of 4.9 m/s. What is the magnitude of the net force acing on the student?

6 0

2 years ago

Read 2 more answers

An upward force act on a proton as it moves with a speed of 2.0 x 10^5 meters/seconds through a magnetic field of 8.5 x 10^2

Gekata [30.6K]

Force on a moving charge is given by formula

$\vec F = q(\vec v \times \vec B)$

here we know that this force will be maximum when velocity is perpendicular to magnetic field

$\vec F = qvB$

here we know that

$v = 2.0 \times 10^5 m/s$

$q = 1.6 \times 10^{-19} C$

$B = 8.5 \times 10^2 T$

now we have

$F = (1.6 \times 10^{-19})(2 \times 10^5)(8.5 \times 10^2)$

$F = 2.72 \times 10^{-11} N$

7 0

3 years ago

How does kinect energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at 6

mote1985 [20]

If it is the same vehicle, then the 60mph vehicle has more kinetic energy since it is moving faster. Therefore, it requires more energy to stop, and if it is the same car with the same beak system, the braking distance of the 30mph car will be significantly shorter than the 60mph car.

8 0

3 years ago