Answer:
t should be 3.57 second
Explanation:
Formula used is v = u+at
In which v is final velocity, u is initial velocity, a is acceleration and t is time.
Substitute each of the info given into the formula and calculate.
49 = 24 + (7)t
t = 3.57s
Answer:
(B) 13.9 m
(C) 1.06 s
Explanation:
Given:
v₀ = 5.2 m/s
y₀ = 12.5 m
(A) The acceleration in free fall is -9.8 m/s².
(B) At maximum height, v = 0 m/s.
v² = v₀² + 2aΔy
(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)
y = 13.9 m
(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.
v = at + v₀
-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s
t = 1.06 s
Answer:




Explanation:
r = Radius of disk = 7.9 cm
N = Number of revolution per minute = 1190 rev/minute
Angular speed is given by

The angular speed is 
r = 2.98 cm
Tangential speed is given by

Tangential speed at the required point is 
Radial acceleration is given by

The radial acceleration is
.
t = Time = 2.06 s
Distance traveled is given by

The total distance a point on the rim moves in the required time is
.
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 = 