The answer depends heavily on what 'objects' you're talking about.
First, find the work done. W = f*d, so 160 N * 1 m = 160 J. Then divide the work by the time to get the power. P = W/t. P = 160 J / 0.5 s = 320 W.
The answer is 320 W. Hope this helps, and have a great day! :)
Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.
Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
Answer:
12.4 m/s²
Explanation:
L = length of the simple pendulum = 53 cm = 0.53 m
n = Number of full swing cycles = 99.0
t = Total time taken = 128 s
T = Time period of the pendulum
g = magnitude of gravitational acceleration on the planet
Time period of the pendulum is given as
T = 1.3 sec
Time period of the pendulum is also given as
g = 12.4 m/s²
Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c
