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A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pulley, which is a disk of radius 9.00 cm , has friction in its axle.What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? (Answer should be in N m)

Answers

Answer 1

Answer:

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

Answer 2

Answer:

Answer:here to earn points

Explanation:

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The surface pressure of the atmosphere is about 14.7 psi (pounds per square inch). How many pounds per square yard does that amount to

Answers

Answer:

14.7 psi is equal to 19051.2 pounds per square yard.

Explanation:

Dimensionally speaking, a square yard equals 1296 square inches. Therefore, we need to multiply the atmospheric pressure by 1296 to obtain its equivalent in pounds per square yard. That is:

$p = 14.7\,(lbf)/(in^(2))* 1296\,(in^(2))/(yd^(2))$

$p = 19051.2\,(lbf)/(yd^(2))$

14.7 psi is equal to 19051.2 pounds per square yard.

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down?

Answers

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 ((v_0)/(v_0-v))$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 ((v_0)/(v_0-v))$

$300 = f_0 ((343)/(343-v))$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 ((v_0)/(v_0-v/2))$

$290 = (343)((v_0)/(343-v/2))$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$((300*343) - 300v)/(290*343-145v) = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

Final answer:

The speed of the train can be determined using the Doppler effect formula.

Explanation:

The question involves the Doppler effect, which is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. In this case, the train whistle's frequency changes from 300 Hz to 290 Hz as the train approaches the station.

To find the speed of the train before and after slowing down, we can use the formula for the Doppler effect:

f' = f((v + v_o)/(v - v_s))

Where:

f' is the observed frequency
f is the source frequency
v is the speed of sound
v_o is the speed of the observer (here it is the train)
v_s is the speed of the source (here it is the speed of sound)

By substituting the given values for observed frequency (290 Hz), source frequency (300 Hz), and the speed of sound (343 m/s), we can solve for the speed of the train before and after slowing down.

Learn more about Doppler effect here:

brainly.com/question/33454469

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The focal length of a concave mirror is 17.5 cm. An object is located 38.5 cm in front of this mirror. How far in front of the mirror is the image located?

Answers

Answer:

Explanation:

object distance u = 38.5 cm ( negative )

focal length f = 17.5 cm ( negative )

mirror formula

1 / v + 1 / u = 1 / f

1 / v - 1 / 38.5 = - 1 / 17.5

1 / v = - 1 / 17.5 + 1 / 38.5

= - 0 .03116

v = - 1 / .03116 = - 32 cm

Image will be formed in front of the mirror at 32 cm distance .

PLEASE HELP IT'S DUE IN LIKE 2 MINUTES

Answers

Answer:

1kg

Explanation:

this box is the smallest and weighs the least. Hope this helps :]

Which are electromagnetic waves? check all that apply.earthquake waves
infrared waves
ocean waves
radio waves
untraviolet waves

Answers

Since electromagnetic waves do not require a medium for their transmission, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

What are electromagnetic waves?

Electromagnetic waves or radiations are waves which occur as a result of the interaction between the electric and magnetic fields.

Electromagnetic waves do not require a material medium for their transmission and as such can travel through a vacuum.

Some examples of electromagnetic waves are radio waves, ultraviolet waves, microwaves, infrared waves etc.

Therefore, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

Learn more about electromagnetic waves at: brainly.com/question/25847009

The electromagnetic waves are:
Radio waves
Ultraviolet waves
And Infrared waves
Hope i helped
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Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g

Answers

Answer:

a = 5.53 g , a = -15g

Explanation:

This is an exercise in kinematics.

a) Let's look for the acceleration

as part of rest v₀ = 0

v = v₀ + a t

a = v / t

a = 282 / 5.2

a = 54.23 m / s²

in relation to the acceleration of gravity

a / g = 54.23 / 9.8

a = 5.53 g

b) let's look at the acceleration to stop

va = 0

0 = v₀ -2 a y

a = vi / y

a = 282/2 1

a = 141 m /s²

a / G = 141 / 9.8

a = -15g

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