Answer:
32 ms
Explanation:
v=32ms.
Explanation:
I will assume that you mean that the acceleration is 8.0 ms2, as 8.0ms is a value for velocity, not acceleration.
Here, we use the formula v=u+at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time. Let's substitute values, then:
v=u+at,
v=0+8.0⋅4.0
v=32,
v=32ms.
Hope it Helps! :D .
Answer: the average speed of the rat from the information given above is 0.7m/s
Explanation:
position is given as
x(t) = pt² + qt
finding the diffencial of x(t) with respect to t, we have
d(x(t))/dt = 2pt + q
we substitute the p = 0.36m/s² and q= -1.10 m/s
d(x(t))/dt = 2(0.36)t + (-1.10)
so, at t= 1s
d(x(t))/dt = 2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s
at t= 4s
d(x(t))/dt = 2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s
To find the average speed,
average speed = (V1 + V2)/ 2
average speed = (1.78 + (-0.38))/2 = 0.7m/s
Answer:
V = 0.9 m/s
Explanation:
The parameters given are:
Initial velocity U = 6.4 m/s
Time t = 0.64s
Height h = 2.05 m
To find the final velocity, let us use third equation of motion
V^2 = U^2 - 2gH
Since the ball is going upward, g will be negative
Substitute all the parameters into the formula
V^2 = 6.4^2 - 2 × 9.8 × 2.05
V^2 = 40.96 - 40.18
V^2 = 0.78
V = sqrt( 0.78)
V = 0.883 m/ s
V = 0.9 m/ s approximately
Answer: The correct answer is : From the period-luminosity relation for Cepheids, he was able to determine the distance to Andromeda and show that it was far outside the Milky Way Galaxy.
Explanation: Hubble's law says that the recession velocity of a galaxy is directly proportional to its distance from us. Hubble measured the distance to the Andromeda galaxy by applying the period-luminosity relationship to Cepheid.
If the scale reads 650N, then the mass of whoever it is standing on the scale is
(weight) / (gravity) = (650N) / (9.8 m/s²) = 66.3 kilograms .
It's not MY mass, even if I'm the one standing on the scale.
If I stand on a scale and it reads 650 N, the scale is broken.
