Answer: 0
Explanation:
75 newtons will push back canceling it out make it 0
a.
The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

where F is the magnitude of the force, theta is the angle between them and d is the distance.
The problen gives the following data:
The magnitude of the force 750 N.
The angle between the force and the displacement which is 25°
The distance, 26 m.
Plugging this in the formula we have:

Therefore the work done is 17673 J.
b)
The power is given by:

the problem states that the time it takes is 6 s. Then:

Therefore the power is 2945.5 W
Answer:
Option C is the untrue statement.
sorry - late reply...just stumbled across tis...hope u can still use it :)
By the mirror equation: 1/di + 1/do = 1/f
<span>
</span>
<span>where di = distance to image = +12cm (+ for real image)</span>
and do = distance to object = +8cm
Substitute and solve for f, the focal length
<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
Answer:
5.5 m/ sec
Explanation:
Because the inclined surface is frictionless so we can assume that total change of energy is zero
i-e ΔE = 0
Or we can say that difference between final and initial energy is zero i-e
Ef- Ei =0
Where,
Ef= final energy at the top of the ramp= KEf+PEf
Ei= Initial energy at the bottom of the ramp=KEi+PEi
So we have
(KEf+PEf)-(KEi+PEi)=0
==>KEf-KEi+PEf-PEi=0 -------------(1)
KEf = mgh = 200×9.8×h
Where h= Sin 22 = h/d= h/4.1
or
0.375×4.1=h
or h= 1.54 m
So, PEf= 200×9.8×1.54=3018.4 j
and KEf= 1/2 m
= 0.5×200×0=0 j
PEi= mgh = 200×9.8×0=0 j
KEi= 1/2 m
=0.5×200×
=100
j
Put these values in eq 1, we get;
0-100
+3018.4-0=0
-100
=-3018.4
==>
= 30.184
==> Vi = 