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Scrat [10]
3 years ago
13

Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 4.5 m long, its mass is 0.4

2 kg, and the force exerted on it by the children is 71 N. (a) What is the linear mass density of the rope (in kg/m)?
(b) What is the speed of the waves on the rope (in m/s)?
Physics
1 answer:
lions [1.4K]3 years ago
7 0

Answer:

density = 0.0933 kg/m

speed = 27.581 m/s

Explanation:

given data

length L = 4.5 m

mass m = 0.42 kg

force  F = 71 N

to find out

mass density and speed

solution

we find linear mass density

linear mass density  = mass / length

put here all value

density = 0.42 / 4.5

density = 0.0933 kg/m

and

speed of wave

speed = √(F/density)

speed = √(0.42/0.933)

speed = 27.581 m/s

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At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m
eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, t_w, is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, d_w, is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking t_w = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × t_w²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × t_w²

t_w = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = t_w ≈ 1.392 s

2) The horizontal distance, d_w, in meters from the edge of the pool to where the diver enters the water is given as follows;

d_w = L + v₀ × t_w = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

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A cyclist acceleration from 0m/s to 8m/s in 3second. what is his acceleration?​
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Two 5.0-cm-diameter conducting spheres are 8.0 m apart, and each carries 0.12 mC. Determine (a) the potential on each sphere, (b
blagie [28]

Answer:

Explanation:

Two spheres 10m apart

Each charge on the sphere is 0.12mC= 0.12×10^-3C

Given that the diameter is

5cm=0.05m

Then, the radius is diameter / 2

r=d/2

r=0.05/2

r=0.025m

Potential is given as

V=kq/r

k=9×10^9Nm2/C2

a. Potential on each sphere surface.

They are going to have the same value since the sphere are identical

At the surface of the sphere,

r= 0.025m

V=kq/r

V=9E9×0.12E-3/0.025

V=4.32E7Volts

V=4.32×10^7Volts

b. The electric field at the surface of each sphere will be the same since the charge are identical,

So, Electric field is given as

E=kq/r^2

At the surface

E=9E9×0.12E-3/0.025^2

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c. The potential mid way between the two sphere

The potential difference due to the first sphere

The sphere are 10m apart then the distance mid way is 5m

Then, the radius of the sphere is 0.025m

Total distance from the center is 5.025m

Then,

V=kq/r

V1=9E9×0.12E-3/5.025

V1=2.149×10^5 Volts

Potential difference due to the second sphere is the same as the first, since both are identical

Total potential is V1 +V2

V=2.149E5+2.149E5

V=4.3E5Volts

Total potential at the middle due to the two sphere is 4.3×10^5Volts

d. The potential difference between the sphere at any point is equal to the potential difference found at c

Therefore the potential difference is

V=4.3×10^5Volts

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3 years ago
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