Answer:
The magnitude of the torque is 263.5 N.
Explanation:
Given that,
Applied force = 31 N
Distance from the axis = 8.5 m
She applies her force perpendicularly to a line drawn from the axis of rotation
So, The angle is 90°
We need to calculate the torque
Using formula of torque
Where, F = force
d = distance
Put the value into the formula
Hence, The magnitude of the torque is 263.5 N.
The object’s resultant angle of motion with the +x-axis after the collision is 47°
<span>From object A:
1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.
Now, we know, tan</span>Ф =
⇒tanФ =
⇒tanФ = 1.088
⇒ Ф =
1.088
= 47.4 ≈ 47
Answer:
<h3>The answer is 80 N</h3>
Explanation:
The force acting on the object can be found by using the formula
where
d is the distance
w is the work done
We have
We have the final answer as
<h3>80 N</h3>
Hope this helps you
Answer:
Explanation:
Given ,
dv / dt = k ( 160 - v )
dv / ( 160 - v ) = kdt
ln ( 160 - v ) = kt + c , where c is a constant
when t = 0 , v = 0
Putting the values , we have
c = ln 160
ln ( 160 - v ) = kt + ln 160
ln ( 160 - v / 160 ) = kt
(160 - v ) / 160 =
1 - v / 160 =
v / 160 = 1 -
v = 160 ( 1 - )
differentiating ,
dv / dt = - 160k
acceleration a = - 160k
given when t = 0 , a = 280
280 = - 160 k
k = - 175
a = - 160 x - 175
a = 28000
when a = 128 t = ?
128 = 28000
= .00457
Answer:
D. 371 N
Explanation:
Friction force equals normal force times the coefficient of friction:
F = Nμ
Since the ground is level, normal force equals weight:
N = mg
Therefore:
F = mgμ
F = (61 kg) (9.8 m/s²) (0.62)
F = 371 N