The spring constant is 147 N/m
Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm
We need to find the spring constant
A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.
The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring
We know that F = kx
300(9.8)= k (0.02)
k = 147.15 N/m
Rounding off to the nearest is 147N/m
The spring constant is 147N/m
Learn more about Hooke's law here
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When the child is moving, he/she has kinetic energy. For just a brief second before they move the other way, the child is not moving, but they have gravitational potential energy.
The child may need a push from time to time because friction with the air causes loss of energy.
The temperature of the lithosphere is around 300<span>°C</span> - 500<span>°<span>C</span></span>
Answer:
a) 17.33 V/m
b) 6308 m/s
Explanation:
We start by using equation of motion
s = ut + 1/2at², where
s = 1.2 cm = 0.012 m
u = 0 m/s
t = 3.8*10^-6 s, so that
0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²
0.012 = 0.5 * a * 1.444*10^-11
a = 0.012 / 7.22*10^-12
a = 1.66*10^9 m/s²
If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where
E = electric field
m = mass of proton
a = acceleration
q = charge of proton
E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19
E = 2.77*10^-18 / 1.6*10^-19
E = 17.33 V/m
Final speed of the proton can be gotten by using
v = u + at
v = 0 + 1.66*10^9 * 3.8*10^-6
v = 6308 m/s
The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
To find the answer, we have to know about the Lorentz transformation.
<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>
It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.
- Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,
- So, to
![V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016](https://tex.z-dn.net/?f=V_x%3D%280.8%2B0.6%29c-%5B%5Cfrac%7B0.6c%2A%280.8c%29%5E2%7D%7Bc%5E2%7D%5D%3D1.016)
find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame. - We have an expression from Lorents transformation for relativistic law of addition of velocities as,
- Substituting values, we get,
Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
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