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3 years ago

What is the magnitude of the net torque on the pulley about the axle?

Physics

1 answer:

jeyben [28]3 years ago

6 0

Answer: Torque= tangential force × radius of the pulley

Explanation:

Pulley is a simple machine which helps us to lift loads by applying the force in a convenient direction. It consists of a grooved wheel which is free to rotate about an axle passing through the center of the pulley.

For a pulley, the torque $T=$ tangential force × radius of the pulley

Pulleys can be used in combination also to achieve a desired output.

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In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark a

Novay_Z [31]

Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

At the 7.43km mark, total distance = 7.43km or 7430m

Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

a = 809.82m / 8016s² = 0.101 m/s²

8 0

2 years ago

2. Place the pairs in order from weakest to strongest gravitational force. (Brainpop Gravity Challenge)

FromTheMoon [43]

Answer:

B, A, C, D

Explanation:

i hope it helps. im just kinda guessing.

good luck!°°°°°⁰⁰⁰⁰⁰₀₀₀₀₀oo00OO

3 0

3 years ago

A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

7 0

2 years ago

Meclanical Energy

miss Akunina [59]

Answer:

a jet flying through the air making a paper airplane

7 0

3 years ago

A 30 kg mass and a 20 kg mass are joined by a light rigid rod and this system is free to rotate in the plane of the page about a

stepan [7]

55 J

Explanation:

Kinetic energy is given as: 0.5MV^2 where M is the mass and V is the speed of rotation. Since the masses are point masses, we calculate the point mass for each mass.

M1 = 30*0.2^2 = 1.2kgm^2

M2 = 20*0.4^2 = 3.2kgm^2

V = 5 rad/s

Calculating using the formula above, we obtain :

0.5(1.2+3.2)5^2 =0.5*4.4*25 = 55 J

3 0

3 years ago

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