Answer:
A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms
2 ).
Mass of the block=5kg
Coeffecient of friction=0.2
external applied force, F=40N
The angle at which the force is applied=30degree
So the horizontal component of force=Fcos30=40×
23 =20 3 N
While the uertical component of the force acting in upward direction=Fsin30=40× 21
=20N
The normal reaction from the surface (N)=mg−Fsin30=50−20=30N
So the ualue of limiting friction=μN=0.2×30=6N
Hence the net horizontal force on the block=Fcos30=μN=20
3
N−6N=28.64N
The horizontal acceleration of the block=
m
Fcos30−μN = 528.64
=5.73m/s 2
Distance is speed x time. Half of the trip is 5.8/2 = 2.9hrs.
640 x 2.9 = 1856mi
580 x 2.9 = 1682mi
1856mi+1682mi=3538mi.
You could also calculate her average speed. This is easy since it was divided in two equal time slices. Average Speed = (640+580)/2 = 610mi/hr
Now 610mi/hr x 5.8hrs = 3538mi
<span>Earth (and hence the observer) moves.</span>
Answer:
1,3,5
Explanation:
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