An athlete runs at a constant velocity of 5.2 m/s. What is the velocity of the athlete relative to the ground?
1 answer:
The relative velocity of the athlete relative to the ground is 5.2 m/s
The given parameters;
constant velocity of the athlete, V = 5.2 m/s
let the velocity of the ground = Vg = 0
The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.
The athlete is the moving object in this question while the ground is stationary.
The relative velocity of the athlete relative to the ground is calculated as follows;
Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s
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The zero net electric field point is at a point that is 0.98 m away from 4.7C charge.If a 14C charge is placed at this point then, force acted on the charge placed at this point is equal to zero.
Explanation:
Let at A both net electric field is zero then
At A ,E1=E2
E1=k*Iq1I / (d+x)^2
E2=k*Iq2I /x^2
Equating both
Answer:
The magnitude of the acceleration of the box is 2.01 m/s².
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following horizontal forces:
Fr = friction force.
Fx = Horizontal component of the applied force, F.
And we have the following vertical forces:
Fy = vertical component of the applied force.
N = normal force exerted on the box.
W = weight of the box.
According to Newton´s second law:
∑F = m · a
Then, in the horizontal direction:
Fx - Fr = m · a
Where "m" is the mass of the box and "a" its acceleration.
Fx can be obtained by trigonometry (see figure):
Fx = F · cos 30°
Fx = 90.0 N · cos 30°
Fr is calculated as follows:
Fr = μ · N
Where μ is the coefficient of friction and N the normal force.
So, we have to find the magnitude of the normal force.
Using Newton´s second law in the vertical direction:
∑F = N + Fy - W = m · a
Notice that the box has no vertical acceleration, then:
N + Fy - W = 0
Solving for N:
N = W - Fy
The weight is calculated as follows:
W = m · g
Where g is the acceleration due to gravity:
W = 20.0 kg · 9.8 m/s² = 196 N
And the vertical component of the applied force can be obtained by trigonometry:
Fy = F · sin 30°
Fy = 90.0 N · sin 30°
The normal force will be:
N = W - Fy = 196 N - 90.0 N · sin 30°
N = 151 N
Now, we can calculate the friction force:
Fr = μ · N
Fr = 0.250 · 151 N
Fr = 37.8 N
And now, we can obtain the acceleration of the box:
Fx - Fr = m · a
(Fx - Fr) / m = a
(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a
a = 2.01 m/s²
The magnitude of the acceleration of the box is 2.01 m/s².
