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Maslowich
3 years ago
14

Assume that the turntable deccelerated during time Δt before reaching the final angular velocity ( Δt is the time interval betwe

en the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, ⟨τ⟩, acting on the bottom disk due to friction with the record?
Physics
1 answer:
Zina [86]3 years ago
6 0

Answer:

See below...

Explanation:

Let’s express ⟨α⟩ in terms of ωi , ωf , and Δt. and torque in terms of It , ωi , ωf , and Δt.

STEP 1.  

The rate of change of angular velocity is Angular acceleration.  

The net change in angular velocity is Average angular acceleration divided by the elapsed time.

⟨α⟩ = ω f −ω i/Δt

STEP 2.

Torque is assumed this way

          dω

   τ = I ----

           dt

.

⟨τ ⟩ = I t (ω f −ω i )/Δt

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Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer
miskamm [114]

Answer:

The cost of energy is $ 0.34.

Explanation:

The energy is the capacity to do work.

The energy is a scalar quantity and its SI unit is Joule.

The commercial unit of energy is kWh.

Cost of 1 kWh energy = $ 0.17

energy loss by standard window is 2 kWh .

So, the cost of lost of energy is

Cost = $ 0.17 x 2 = $ 0.34

3 0
3 years ago
A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri
exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

where

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at halfway

K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

E=E_e = \frac{1}{2}kx^2

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

E=E_e = 14.3 J

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

E=U_f = mg h_f

where h_f is the maximum height reached. Solving for this quantity, we find

h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

5 0
3 years ago
A bond between a positively-charged particle and a negatively-charged particle is a(n) _______ bond.
weeeeeb [17]
The answer is an ionic bond.
6 0
3 years ago
How much work is done when an engine generates 400 Watts of power in 25 seconds?
IceJOKER [234]

Answer:

10000 J or 10 KJ

Explanation:

power = workdone/time taken

400 = workdone/25

workdone = 400 * 25

=10000 J

8 0
3 years ago
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kvasek [131]
D.6.22N. because .42kg * 14.8m/s=6.22 N[meaning newtons}.
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3 years ago
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