Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?
a-velocity
b-mass
c-momentum
d-direction
Answer:
b. Mass
Explanation:
This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.
As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.
Therefore, the correct answer is B. Mass.
Answer:
3.7 A
Explanation:
Parameters given:
Magnetic field strength, B = 5 * 10^(-5) T
Distance of magnetic field from wire, r = 1.5 cm = 0.015 m
The magnetic field, B, due to a current, I, flowing a wire is given as:
B = (μ₀*I) / 2πr
Where μ₀ = permeability of free space
To get the current, I, we make I the subject of the formula:
I = (2πr * B) / μ₀
I = (2 * 3.142 * 5 * 10^(-5)) / (1.25663706 × 10^(-6))
I = 3.7 A
Answer:
133.8 N
Explanation:
Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2
Therefore, the weight of the dog on this planet would be:
Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N
Answer:
sorry about the other person but its b
Explanation:
Explanation:
Formula which holds true for a leans with radii 
and 
and index refraction n is given as follows.
![\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%28n%20-%201%29%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
Since, the lens is immersed in liquid with index of refraction 
. Therefore, focal length obeys the following.
![\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf_%7B1%7D%7D%20%3D%20%5Cfrac%7Bn%20-%20n_%7B1%7D%7D%7Bn_%7B1%7D%7D%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
![\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%28n%20-%201%29%7D%20%3D%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
and, 
or, 
= 32.4 cm
Using thin lens equation, we will find the focal length as follows.
Hence, image distance can be calculated as follows.
= 47.9 cm
Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.
