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Answer:
A) v = 40 m / s, B) v_average = 20 m / s
Explanation:
For this exercise we will use the kinematics relations
A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero
v = vo + a t
v = 0 + 8 5
v = 40 m / s
B) the average velocity can be found with the relation
v_average = vf + vo / 2
v-average = 0+ 40/2
v_average = 20 m / s
Answer:
distance = 6.1022 x10^16[m]
Explanation:
To solve this problem we must use the formula of the average speed which relates distance to time, so we have
v = distance / time
where:
v = velocity = 3 x 10^8 [m/s]
distance = x [meters]
time = 6.45 [light years]
Now we have to convert from light-years to seconds in order to get the distance in meters.
![t = 6.45 [light-years]*365[\frac{days}{1light-year}]*24[\frac{hr}{1day}] *60[\frac{min}{1hr}]*60[\frac{seg}{1min} ] =203407200 [s]](https://tex.z-dn.net/?f=t%20%3D%206.45%20%5Blight-years%5D%2A365%5B%5Cfrac%7Bdays%7D%7B1light-year%7D%5D%2A24%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%20%2A60%5B%5Cfrac%7Bmin%7D%7B1hr%7D%5D%2A60%5B%5Cfrac%7Bseg%7D%7B1min%7D%20%5D%20%3D203407200%20%5Bs%5D)
Now using the formula:
distance = v * time
distance = (3*10^8)*203407200
distance = 6.1022 x10^16[m]
Answer:
-2.33 m/s²
Explanation:
The computation of the skateboarder’s acceleration is shown below;
Acceleration means the change in velocity per unit with respect to time.
In the given case, the initial velocity is 7 m/s.
As in the question it is mentioned that it comes to a stop, so the final velocity would be zero.
And, The time elapsed is 3 seconds.
Now the following equation should be used
a = (v,final - v,initial) ÷ t
= (0 - 7)/3
= -2.33 m/s²
Answer:
harder
heavier objects resist change more than lighter objects.
