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baherus [9]
3 years ago
12

A string is wound tightly around a fixed pulley having a radius of 5.0 cm. as the string is pulled, the pulley rotates without a

ny slipping of the string. what is the angular speed of the pulley when the string is moving at 5.0 m/s?
Physics
2 answers:
lubasha [3.4K]3 years ago
8 0
The angular speed can be solve using the formula:
w = v / r
where w is the angular speed
v is the linear velocity
r is the radius of the object

w = ( 5 m / s ) / ( 5 cm ) ( 1 m / 100 cm )
w = 100 per second
Kruka [31]3 years ago
4 0

Answer:

Angular speed of the pulley = 100 rad/s

Explanation:

Linear speed = Angular speed x Radius

    v = ωr

Given that the string is moving at 5.0 m/s

That is v = 5 m/s

The string is wound tightly around a fixed pulley having a radius of 5.0 cm,

           r = 5 cm = 0.05 m

Linear speed = Angular speed x Radius

5 = ω x 0.05

ω = 100 rad/s

Angular speed of the pulley = 100 rad/s

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lions [1.4K]

Answer: Choose the normal force acting between the object and the ground. Let's assume a normal force of 250 N.

Determine the friction coefficient.

Multiply these values by each other: 250 N * 0.13 = 32.5 N .

You just found the force of friction!

Explanation:

8 0
2 years ago
A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20
Klio2033 [76]

Answer:

\boxed{{\boxed{\blue{ 12.5}}}}

Explanation:

Given, for girl : Weight or force;

\rm \: F_1 = 50 \: kgf

Area of both heels;

\rm \: A_1 =  \; 2 ×1 \;  cm^2 = 2  \: cm^2

\rm \: Pressure \:  P_1  =  \cfrac{F_1}{ A_1 }  =  \dfrac{50 \: kgf}{2 \: cm {}^{2} }  = 25 \: kgf \: cm {}^{ - 1}

For elephant, Weight = Force \rm F_2 = 2000 kg•f

Area of 4 feet;

\rm \: A_2  = \; 4 \times 250 \;  cm^2 = 1000 \:  cm^2

\rm \: Pressure \:  P_2 = {F_2}/{A_2} \;  = \cfrac{2 \cancel{0 00 }\:  kgf}{1 \cancel{000} \: cm^2} =  2 \: kgf \: { \:cm}^{- 1}

Now;

\rm  = \dfrac{Pressure \:  Exerted  \: by  \: the \:  Girl}{Pressure  \: exerted  \: by \:  the  \: elephant}

=  \rm \: P_1/P_2

\implies    \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } =  \rm\cfrac{25 \:  \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0
2 years ago
Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid
Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0
3 years ago
A cord of mass 0.65 kg is stretched between two supports 8.0 m apart. If the tension in the cord is 120 N, how long will it take
DedPeter [7]

The time taken by the pulse to travel from one support to the other is 0.208 s.

<h3>Given:</h3>

The mass of the cord is m = 0.65 kg.

The distance between the supports is, d = 8.0 m.

The tension in the cord is T = 120 N.

The time taken by the pulse to travel from one support to the other is given as,

v=\frac{d}{t}

t=\frac{d}{v}

Here, v is the linear velocity of a pulse. Its value is,

v=\sqrt{\frac{T d }{m} }

v=\sqrt{\frac{120 * 8}{0.65} }

v= 38.43 m/s

Then,

t=\frac{8}{38.43}

t=0.208 s

Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.

Learn more about tension here:

brainly.com/question/24994188

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7 0
1 year ago
A moving roller coaster speeds up with constant acceleration for 2.3\,\text{s}2.3s2, point, 3, start text, s, end text until it
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Answer:

Δx=(v+v0/2)t

Explanation:

We can figure out which kinematic formula to use by choosing the formula that includes the known variables, plus the target unknown.

In this problem, the target unknown is the initial velocity v_0v  

0

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v, start subscript, 0, end subscript of the roller coaster.

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