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katrin2010 [14]
4 years ago
5

A painted tooth on a spinning gear has angular acceleration α=(20−t)rad/s2, where t is in s. Its initial angular velocity, at t

= 0 s, is 320 rpm.
Physics
1 answer:
oksano4ka [1.4K]4 years ago
4 0

Answer:

I think the question will be to find the angular speed at any time t

Since we are given the initial conditions of the angular speed

Explanation:

Given that,

α=(20−t)rad/s2

Give the initial condition

At t=0 w=320rpm

So modeling a differential equation

α=dw/dt

Therefore

dw/dt=(20−t)

Applying variable separation

dw=(20-t)dt

Integrate Both sides

∫dw=∫(20-t)dt

w = 20t-t²/2 + C

C is constant of integration

Using initial conditions

At t=0 w=320rpm

320=0-0+C

C=320rpm

Therefore the equation becomes

W=20t - t²/2 + 320

This is the model angular speed at any time t

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A cord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and of mass 8.00 kg. A steady horizontal pull of 34.
disa [49]

Answer:

A. α = 94.4 rad/s

B. a = 28.32 m/s

C. N = 34N

D. α = 94.4 rad/s

    a = 28.32 m/s

     N =  44.4  N

Explanation:

part A:

using:

∑T = Iα

where T is the torque, I is the moment of inertia and α is the angular momentum.

firt we will find the moment of inertia I as:

I = \frac{1}{2}MR^2

Where M is the mass and R is the radius of the wheel, then:

I = \frac{1}{2}(8 kg)(0.3 m)^2

I = 0.36 kg*m^2

Replacing on the initial equation and solving for α, we get::

∑T = Iα

Fr = Iα

34 N =  0.36α

α = 94.4 rad/s

part B

we need to use this equation :

a = αr

where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:

a = (94.4)(0.3 m)

a = 28.32 m/s

part C

Using the laws of newton, we know that:

N = T

where N is the force that the axle exerts on the wheel part and T is the tension of the cord

so:

N = 34N

part D

The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:

-Mg + N +T = 0

Then, solving for N, we get:

N = -T+Mg

N = -34 + (8 kg)(9.8)

N =  44.4  N

6 0
3 years ago
How does an inclined plane trade force for distance??
liq [111]
The sloping surface of the inclined plane<span> supports part of the weight of the object as it moves up the slope. As a result, it takes less </span>force<span> to move the object uphill. The </span>trade<span>-off </span>is<span> that the object must be moved over a greater </span>distance<span> than if it were moved straight up to the higher elevation.</span>
8 0
4 years ago
Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while
Vinvika [58]

Answer:

Explanation:

Given

Time taken to reach ground is t=2.7\ s

Malda initial velocity u=95\ cm/s

Let h be the height of Cliff

using h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

h=0+\frac{1}{2}(9.8)(2.7)^2

h=35.72\ m

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

R=u\times t

R=0.95\times 2.7=2.43\ m    

7 0
3 years ago
A physical science test book has a mass of 2.2 kg what is the weight on the earth
timofeeve [1]

Answer:

21.582 Newtons

Explanation:

The weight of an object on Earth is equal to W=9.81M where M is the mass of the object and the acceleration due to gravity is 9.81 m/s^2. Therefore, the weight of the book on the earth is W=9.81(2.2)=21.582N, or 21.582 Newtons

3 0
3 years ago
Find the magnitude of the horizontal force, in N, required to push an object a distance of 2.00 meters if the force did 0.0430 J
Sergio039 [100]
Your force is 0.0215N
W=fxd and your're looking for F so you make the equation equal to F
f=w/d so 0.0430/2=0.2215N
5 0
3 years ago
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