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Neporo4naja [7]
3 years ago
7

A yo-yo has a string that is 0.95 m in length. What is the period of oscillation if the yo-yo is allowed to swing back and forth

at the end of its string?
Physics
1 answer:
yulyashka [42]3 years ago
5 0

Answer:

Explanation:

As we know the , equation of time period for simple pendulum ,

T = 2*pi*\sqrt{l/g}

hence putting values we get ,

the solution is in picture ,

please  

Brain-list it or support me at my U-Tube channel " ZK SOFT&GAMING " I will be thankful

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Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per
Schach [20]

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90^{\circ} to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

7 0
3 years ago
A note of frequency 2000Hz has a velocity of 400 ms! What is the wavelength of the note?​
Rudik [331]

Answer:

Explanation:

frequency=c/λ

c=400m/s

putting values

2000=400/λ

λ=0.2m

8 0
3 years ago
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A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
Can you answer these 2 ?
vredina [299]

Answer:

1. 38,500

2. 308,000

Explanation:

This would require a calculator. To find momentum, you multiply mass and velocity. You always want your mass to be measure in kilograms, but that is irrelevant in this question because they already are, it is just something to remember.

4 0
2 years ago
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