A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block
1 answer:
Answer:
+ 24 N
Explanation:
the computation is shown below:
Given that
Mass of the block = m = 0.7 kg
Sprint constant = k = 160 N / m
x = 0.15m
Now the force on the block is
F = kx
= (160) (0.15)
= 24 N
As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction
Therefore the third option is correct
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Answer:
The velocity of the plane at take off is 160 m/s.
The distance travel by the plane in that time is 3200 meter.
Explanation:
Given:
Acceleration, a = 4 m/s²
Time, t = 40 s
u = 0 i .e initial velocity
To Find:
velocity , v = ?
distance , s =?
Solution:
we have first Kinematic equation
v = u + at
∴ v = 0 + 4×40
∴ v = 160 m/s
Now by Third Kinematic equation
∴ s = 0 + 0.5 × 4× 40²
∴ s = 3200 meter
Answer:
Explanation:
Considering an object that moving about in a circular path, the equation for such centripetal force can be computed as:
The model for the person can be seen in the diagram attached below.
So, along the horizontal axis, the net force that is exerted on the person is:
Dividing both sides by "m"; we have :
Making "v" the subject of the formula: we have:
So, when = 0; the velocity is maximum
∴
Hence; the maximum walking speed for the person is
