Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151
Answer:
The rate at which radar must rotate is 0.335 rad/s.
Explanation:
Given that,
Velocity = 65 m/h = 29.0576 m/s
Angle = 15°
Suppose, the radius given by

We need to calculate the rate at which radar must rotate
Using formula of linear velocity


Where, v = velocity
r = radius
Put the value into the formula
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
Hence, The rate at which radar must rotate is 0.335 rad/s.
The roller coaster is moving in a circular path, so the force that must be computed is the centripetal force. The centripetal force is the force acting on an object undergoing circular motion and is directed towards the center of the circular path. This is computed using:
F = mv²/r
F = (500 * 18²) / 12
F = 13,500 N
Now, at the bottom of the track, the track is also supporting the weight of the car and its passengers, which is:
W = mg
W = 500 * 9.81
W = 4,905 N
The total reactive force exerted by the track to counter the centripetal force and the weight of the car is:
F = 13,500 + 4,905
F = 18,405 Newtons
They best represent a wave with zero energy and zero amplitude.
There are no measurements shown in a table that accompanies
this question having any amplitude or energy greater than zero.
Remember what acceleration is? It's how fast the speed is changing over a period of time.

So in dt = 2 secs, she sped up from 6 to 10 m/s and so dv = (10-6) m/s.
Now that you know dv and dt, you can calculate the acceleration a!