5. A body travels in a circle at a constant speed, it

2 answers:

7 0

No acceleration as the term acceleration describes change in velocity so if its speed is same there will be no acceleration

nata0808 [166]2 years ago

5 0

A the same acceleration

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A tiger leaps with an initial velocity of 35.0 km/hr at an angle of 13.0ᶿ with respect to the horizontal. What are the component

Lorico [155]

Answer:

Vx = 35 x cos(13deg)

Vy = 35 x sin(13deg) - gt

(g is acceleration due to gravity =~9.8 meter/second^2, t is time in second)

Explanation:

The tiger leaps up, then x and y component of its velocity are:

Vx = Vo x cos(alpha)

Vy = Vo x sin(alpha) - gt

(Vo is tiger's initial velocity, alpha is angle between its leaping direction and horizontal plane)

Hope this helps!

3 0

2 years ago

Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger t

podryga [215]

Answer:

8F_i = 3F_f

Explanation:

When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.

Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.

Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.

The electrostatic force, Fi, in the initial configuration can be calculated as follows.

$F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f$