Explanation:
Speed is distance over time.
500 km / 2 hr = 250 km/hr
Velocity is speed and direction.
250 km/hr north
Answer:
Δω = -5.4 rad/s
αav = -3.6 rad/s²
Explanation:
<u>Given</u>:
Initial angular velocity = ωi = 2.70 rad/s
Final angular velocity = ωf = -2.70 rad/s (negative sign is
due to the movement in opposite direction)
Change in time period = Δt = 1.50 s
<u>Required</u>:
Change in angular velocity = Δω = ?
Average angular acceleration = αav = ?
<u>Solution</u>:
<u>Angular velocity (Δω):</u>
Δω = ωf - ωi
Δω = -2.70 - 2.70
Δω = -5.4 rad/s.
<u> Average angular acceleration (αav):</u>
αav = Δω/Δt
αav = -5.4/1.50
αav = -3.6 rad/s²
Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.
Work needed: 720 J
Explanation:
The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

where
k is the spring constant
x is the stretching of the spring from the equilibrium position
In this problem, we have
E = 90 J (work done to stretch the spring)
x = 0.2 m (stretching)
Therefore, the spring constant is

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of
x = 0.2 + 0.4 = 0.6 m
Substituting,

Therefore, the additional work needed is

Learn more about work:
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Answer:
554.27N
Explanation:
(a) The max frictional force exerted horizontally on the crate and the floor is,
Substitute the values,
μs=0.5
mass=113kg
g=9.81m/s
Ff=μsN
=μsmg
=(0.5 x 113 x 9.81)
Ff=554.27N
1. Cenozoic ERA
2. Mesozoic ERA
3. Paleozoic ERA