Where are the following sketches?
<span>Use the kinematic equation vf^2 = vi^2 + 2ad where;
vf = ?
vi = 0 m/s
a = 9.8 m/s^2
d1 = 10 m
d2 = 25 m
final velocity at the ground (d1): vf = sqrt(2)(9.8)(10) = 14 m/s
final velocity to the bottom of the cliff (d2): vf = sqrt(2)(9.8)(25) = 22.14 m/s
</span>
This question deals with the law of conservation of momentum, which basically says that the total momentum in a system must stay the same, provided there are no outside forces. Since you were given the mass and velocity of the two objects you can find the momentum (p=mv) of each and then add them together to find the total momentum of the system before they collide. This total momentum must be the same after they collide. Since you have the mass and velocity of one of the objects after the collision you can find the its momentum after. Subtract this from the the system total and you will have the momentum of the other object after the collision. Now that you know the momentum of the other object you can find its velocity using p=mv and its mass from before.
Be careful with the velocities. They are vectors, so direction matters. Typically moving to the right is positive (+) and moving to the left is negative (-). It is not clear from your question which direction the objects are moving before and after the collision.
Answer:
d = 0.645 m <em>(assuming a radius of the ball bearing of 3 mm)</em>
Explanation:
<u>The given information is:</u>
- <em>The distance from the center of the sun to the center of the earth is 1.496x10¹¹m =
</em> - <em>The radius of the sun is 6.96x10⁸m =
</em>
<u>We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm =
</u>.
First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth,
:
[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m
I hope it helps you!
Answer:
(a) 160000 kV/m
(b) 1336 keV
Explanation:
(a) magnetic filed, B = 10 T
energy of electron, E = 740 eV
mass of electron, m = 9.1 x 10^-31 kg
Let v be the velocity of electron.
E = 1/2 mv^2
740 x 1.6 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2
v = 1.6 x 10^7 m/s
v = E / B
E = v x B = 1.6 x 10^7 x 10 = 16 x 10^7 V/m
E = 160000 kV/m
(b) E = 16 x 10^7 V/m
B = 10 T
Let v be the velocity of protons.
v = E / B = 16 x 10^7 / 10 = 1.6 x 10^7 m/s
Kinetic energy of proton, E = 1/2 mv^2
= 0.5 x 1.67 x 10^-27 x 1.6 x 1.6 x 10^14
= 2.14 x 10^-13 J = 1336000 eV = 1336 keV