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Rashid [163]
2 years ago
11

What is the angular speed of a body vibratingat 50 cycles per second:​

Physics
1 answer:
tiny-mole [99]2 years ago
6 0

Answer:

100Πrad/s

Explanation:

f = 50cycles/sec = 50HZ

w = 2Πf = 2Π(50) = 100Π

w = 100Πrad/s

Hope this helps...

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What kind of energy does a rubber band have when it is stretched?
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Elastic potential energy. When you stretch a rubber band it has the "potential" to do work, to fly in a given direction. In doing so it changes it's elastic potential energy to kinetic energy.
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8 0
1 year ago
The poles of the Earth's magnetic field line up with the geographic poles of the Earth.
Varvara68 [4.7K]

Answer: False

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3 years ago
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A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top
Advocard [28]

Answer:

a)  Em₀ = 42.96 104 J , b)   W_{fr} = -2.49 105 J , c)  vf = 3.75 m / s

Explanation:

The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

        Em = K + U

a) Let's look for the initial mechanical energy

      Em₀ = K + U

      Em₀ = ½ m v2 + mg and

      Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

      Em₀ = 36 104 + 6.96 104

      Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

    W_{fr} = Em₂ -Em₀

     Em₂ = K + U

     Em₂ = ½ m v₂² + m g y₂

     Em₂ = ½ 50 85 2 + 50 9.8 427

     Em₂ = 180.625 + 2.09 105

     Em₂ = 1,806 105 J

     W_{fr} = Em₂ -Em₀

     W_{fr} = 1,806 105 - 4,296 105

     W_{fr} = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

       W_{fr} = ΔEm

       W_{fr} = Emf - Emo

       -1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

       ½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

       ½ 50.0 vf² = 0.561

       vf = √ 0.561 25

      vf = 3.75 m / s

6 0
3 years ago
If μs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. calculate this
fiasKO [112]

weight = mg acts downwards <span>
normal force = N acts upwards. 
and force F acts at an angle θ below the horizontal. 
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so that, Frictional force, f=us*N acts towards the left 

Now we balance the forces along x and y directions: 
y direction: N = mg + F sinΘ 
x direction: us * N = F cosΘ 

We let the value of µs be equal to a value such that any F will not be able to move the crate. Then, if we increase F by an amount F', then the force pushing the crate towards the right also increases by F' cosΘ. Additionally, the frictional force f must raise by exactly this amount. 
Since f can’t exceed us*N, so the normal force must increase by F' cosΘ/us. 
Also, from the y direction equation, the normal force exceeds by F' sin Θ. 

<span>These two values must be the same, therefore:
<span>us = cot θ</span></span></span>

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