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3 years ago

A ball on a string makes 25.0 revolutions in 9.37 s, in a circle radius 0.450 m. What is it’s velocity?

Physics

1 answer:

Fudgin [204]3 years ago

8 0

Answer:

7.54 m/s

Explanation:

C = 2πr

C = 2π (0.450 m)

C = 2.827 m

v = d / t

v = 25.0 (2.827 m) / (9.37 s)

v = 7.54 m/s

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2. A 2kg body attached to a spring undergoes a SHM of amplitude 0.4m and period

Rudik [331]

Answer:

a) 12.8 N

b) 3.2 m/s²

Explanation:

I'm guessing the period is 0.5π s.

Period of a spring in simple harmonic motion is:

T = 2π √(m/k)

Given T = 0.5π and m = 2 kg:

0.5π = 2π √(2/k)

0.25 = √(2/k)

0.0625 = 2/k

k = 32

The spring constant is 32 N/m, and the maximum displacement is 0.4 m. The maximum force can be found with Hooke's law:

F = kx

F = (32 N/m) (0.4 m)

F = 12.8 N

The acceleration can be found with Newton's second law:

∑F = ma

kx = ma

(32 N/m) (0.2 m) = (2 kg) a

a = 3.2 m/s²

8 0

3 years ago

If acceleration of a particle at any time is given by a=2t+5 the velocity after 5 seconds, if it starts from rest is?

kow [346]

The velocity after 5 seconds is 50 m/s.

3 0

3 years ago

Calculate the velocity of the car between 20 and 30 sec if it is 5/30-4/20

Bezzdna [24]

Answer:

here you go 4 m/s^2......tadaa

5 0

3 years ago

A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh

zheka24 [161]

Answer:

Magnetic force, F = 0.24 N

Explanation:

It is given that,

Current flowing in the wire, I = 4 A

Length of the wire, L = 20 cm = 0.2 m

Magnetic field, B = 0.6 T

Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

$F=ILB\ sin\theta$

$F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)$

F = 0.24 N

So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.

7 0

4 years ago

Read 2 more answers

While camping in Denali National Park in Alaska, a wise camper hangs his pack of food from a rope tied between two trees, to kee

Crank

The tension in each of the ropes is 625 N.

Draw a free body diagram for the bag of food as shown in the attached diagram. Since the bag hangs from the midpoint of the rope, the rope makes equal angles θ with the horizontal. The tensions <em>T</em> in both the ropes are also equal.

Resolve the tension T in the ropes into horizontal and vertical components T cosθ and T sinθ respectively, as shown in the figure. At equilibrium,

$2T sin\theta = 50 N$ ......(1)

Calculate the value of sinθ using the right angled triangles from the diagram.

$sin\theta =\frac{0.06 m}{1.5 m} =0.04$

Substitute the value of sinθ in equation (1) and simplify to obtain T.

$2T sin\theta = 50 N\\ T= \frac{50 N}{2 sin\theta}=\frac{50 N}{2*0.06} = 625 N$

Thus the tension in the rope is 625 N.

4 0

4 years ago