Answer:
26.6 m/s
Explanation:
Given:
Δy = 2.1 m
t = 5.35 s
a = -9.8 m/s²
Find: v₀
Δy = v₀ t + ½ at²
(2.1 m) = v₀ (5.35 s) + ½ (-9.8 m/s²) (5.35 s)²
v₀ = 26.6 m/s
Answer:

Explanation:
Please find the image for the question as attached file.
Solution -
Given -
First of all we will calculate the velocity at point C,
As per newton's third law of motion-

Substituting the given values in above equation, we get -

Now we will determine the radius of curvature for the curve shown in the attached image

Differentiating on both the sides, we get -
meter
Acceleration on curved path

Final acceleration

Explanation:
Given that,
Radius of the circular loop, r = 3.5 cm = 0.035 m
(a) During a 0.12-s time interval, the magnitude of the field increases uniformly from 0.2 T to 0.5 T. Due to the change in the magnetic field, an emf will induced in it. The magnitude of induced emf is given by :

So, the magnitude of the emf induced in the loop during the time interval is
.
(b) The negative sign shows that the direction of induced emf in the loop is in anitclockwise direction.
Answer:
counter question if you get out the shower clean then how does your towel get dirty?