Answer:
From certain assumptions that the walking speed is 2 m/s, and the stop time is 0.1 s the acceleration would be -20 m/s
Explanation:
Using the average acceleration formula:
where 
and 
are the changes in the speed and time respectively.
We have by assuming that the walking speed is 2 m/s and the stop time is 0.1s which is equal to the change in time during the stopping.
, where 
are the initial speed and final speed respectively, and 
Plugging the previous in the average acceleration formula we get
where the minus sign indicates an acceleration in the opposite direction of the motion (or in other word opposite to the speed's direction).
Answer:
The spring constant of this spring is 200 N/m.
Explanation:
Given:
Original unstretched length of the spring (x₀) = 10 cm =0.10 m [1 cm =0.01 m]
Stretched length of the spring (x₁) = 18 cm = 0.18 cm
Force acting on the spring (F) = 16 N
Spring constant of the spring (k) = ?
First let us find the change in length of the spring or the elongation caused in the spring due to the applied force.
So, Change in length = Final length - Initial length
Now, restoring force acting on the spring is directly related to its elongation or compression as:
Rewriting in terms of 'k', we get:
Now, plug in the given values and solve for 'k'. This gives,
Therefore, the spring constant of this spring is 200 N/m.
It’s like the force put against something or someone✨
The answer is a White Dwarf.
Answer:
(a) 1767.43 N
(b) 182.45 N
Explanation:
Radius of earth, R = 6450 km
Weight of person, W = 7070 N
mass of person, m = W / g = 7070 / 9.8 = 721.4 kg
(a) h = 6450 km
The value of acceleration due to gravity on height is given by
g' = g / 4 = 9.8 / 4 = 2.45 m/s^2
The weight of the person at such height is
W' = m x g' = 721.4 x 2.45
W' = 1767.43 N
(b) h = 33700 km
The value of acceleration due to gravity on height is given by
g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2
The weight of the person at such height is
W' = m x g'
W' = 721.4 x 0.253
W' = 182.45 N
