we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m
Answer:
Here we do not have the vector, but I will try to give a kinda general solution to this type of problem.
If the vector is written as (a, b, c) we have that the force in the x-axis is of a Newtons, in the y-axis is of b Newtons, and in the z-axis is of c Newtons.
Then, we can calculate the total magnitude of this force as:
F = √( a^2 + b^2 + c^2)
wich gives us the total magnitude of the force, but not a direction or anything like that, this is just a scalar.
Answer:
4.408 m/s, 4.102 m/s, 4.026 m/s
Explanation:
The question is incomplete. The text of the original question states:
A race car moves such that its position fits the relationship
:

where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.
We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:

And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:

The answer is
Neither the speed of light in air is going to stay the same no matter what wavelength or frequency
Answer:
option A
Explanation:
Escape velocity of the planet

now, it is given that
Second Planet
R₂ = 2R and M₂ = 2M
now escape velocity of the second planet
now,

on solving


escape velocity of the second planet is equal to first.
The correct answer is option A