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docker41 [41]
3 years ago
8

In a tape recorder, the tape is pulled past the read-and-write heads at a constant speed by the drive mechanism. Consider the re

el from which the tape is pulled -- as the tape is pulled off it, the radius of the roll of remaining tape decreases. How does the torque on this reel change with time
Physics
1 answer:
Irina-Kira [14]3 years ago
6 0

Answer:

Torque decreases .

Explanation:

The tape is pulled at constant speed , speed v is constant , so there is

v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .

Let it be α . Let I be moment of inertia of reel .

Reel is in the form of disc

I = 1/2 m r²

I x α = torque

1/2 m r² x α =  torque

As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .

You might be interested in
A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity
nevsk [136]
We can use the kinematic equation
(v_f)^2 = (v_i)^2 + 2*a*d
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance

we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s   [if you are rounding this by significant figures]
8 0
3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of ?5.20 rad/s2. during a 3.80-s time inte
ddd [48]

<span>We can answer this using the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²     -----> 1</span></span>

ω² = ω₀² + 2αθ            -----> 2

Where:

θ = final angular displacement = 70.4 rad

θ₀ = initial angular displacement = 0

ω₀ = initial angular speed

ω = final angular speed

t = time = 3.80 s

α = angular acceleration = -5.20 rad/s^2

Substituting the values into equation 1:<span>
70.4 = 0 + ω₀(3.80) + ½(-5.20)(3.80)² </span><span>

ω₀ = (70.4 + 37.544) / 3.80 </span><span>

ω₀ = 28.406 rad/s </span><span>


Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4 


ω = 8.65 rad/s 


</span>

5 0
3 years ago
A force of 6.0 N is applied horizontally to a 3.0 kg crate initially at rest on a horizontal frictionless surface. After the cra
monitta
Your answer to your question is 6.9 m/s
5 0
3 years ago
A man lifts various loads with the same lever. The distance of the applied force from the fulcrum is 2.00 m and the distance fro
lakkis [162]
In your question where as ask the man lift various loads with the same lever and the distance of the applied force from the fulcrum is 2m and the distance from the fulcorm to load is 0.5m.So the following are the answers to your questions:
a.mechanical advantage of the lever is 1
b..<span>ideal mechanical advantage of the lever is 4</span>
c.4
7 0
3 years ago
Read 2 more answers
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