Answer:
The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.
Explanation:
Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

Dividing the second equation by the first one, we obtain:

And, since
, then:

It means that the velocity at the bottom of the ramp is 1.81m/s.
We could use this data, plus any of the two initial equations, to determine the acceleration:

So the acceleration is 3.30m/s^2.
Answer:

Explanation:
According to the free body diagram, in this case, we have:

Recall that the force of friction is given by:

Replacing and solving for the coefficient of kinetic friction:

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

Finally, we calculate
:

Answer:
5.09 x 10⁵ Nm²/C
Explanation:
The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e
φ = E A
From the question;
E = (8.0j + 2.0k) ✕ 10³ N/C
r = radius of the circular area = 9.0m
A = area of a circle = π r² [Take π = 3.142]
A = 3.142 x 9² = 254.502m²
Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.
Therefore;
φ = (2.0) x 10³ x 254.502
φ = 5.09 x 10⁵ Nm²/C
The electric flux is 5.09 x 10⁵ Nm²/C
Answer:
21/5 seconds or 4.2 seconds
Explanation:
Distance=Speed*Time
21=5 *Time
21/5=Time
4.2=time