Answer:
∴ [T]=[WF−1V−1]
Hope this answer is right!!
Answer:
Force that acted on the body was F = 13 N
Explanation:
If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s
When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:

which in our case becomes;

and we can solve for the acceleration as:
a = 10/10 m/s^2 = 1 m/s^2
Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:
F = 13 kg * 1 m/s^2 = 13 N
Answer:
-2.26×10^-4 radians
Explanation:
The solution involves a right angle triangle
Length is z while the horizontal is the height x
X^2+ 100^2=z^2
Taking the derivatives
2x(dx/dt)=Z^2(dz/dt)
Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11
dz/dt= 1100sqrt3/200 = 9.53
Sin a= 100/a
Taking derivatives in terms of t
Cos a(da/dt)=100/z^2 dz/dt
a= 30°
Cos (30°)da/dt= (-100/40000×9.5)
a= -2.26×10^-4radians
Answer:
there are 7 significant figures
Explanation:
15.33879+15.555
=30.89379
there are 7 significant figures
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Answer:10.4 times of initial velocity
Explanation:
Given
Diameter reduced by 69 %
it approaches with velocity 
suppose its velocity is v during blocked passage
suppose d is the initial diameter and
diameter is



As flow is constant


