A block slides forward 4.73 m while being pulled back by a 39.7 N force at 142° from the direction of motion. How much work does
1 answer:
Answer:
work done by the force, W = -147.97 J
Given:
displacement of the box, s = 4.73 m
angle between force and displacement,
Force, F = 39.7 N
Solution:
Work done by the force can be expressed as the dot product of Force, F and displacement 's' is given by:
W = Fscos
W = 39.7
W = -147.97 J
where negative sign indicates that work is done on the block
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Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude
ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
Answer: The degree of the first term.
Explanation:
The function:
The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.
Let the left hand side be donated by -x.
Then,
Hence, the correct option is the degree of the first term indicates the left and right end points of the function.
Answer:
300 m
Explanation:
The train accelerate from the rest so u = 0 m/sec
Final speed that is v = 80 m/sec
Time t = 30 sec
The distance traveled by first plane = 1200 m
We know the equation of motion where s is distance a is acceleration and u is initial velocity
Using this equation for first plane
As the acceleration is same for both the plane so a for second plane will be 2.67
The another equation of motion is using this equation for second plane
s = 300 m