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Salsk061 [2.6K]
3 years ago
14

The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. Assume that the rope and pu

lley are massless and that there is no friction in the pulley. If the masses have the values m1=21.1 kg and m2=12.9 kg, find the magnitude of their acceleration ???? and the tension T in the rope. Use ????=9.81 m/s2.

Physics
1 answer:
Kisachek [45]3 years ago
6 0

Answer:

a=2.36\ m/s^2

T=157.06 N

Explanation:

Given that

Mass of first block = 21.1 kg

Mass of second block = 12.9 kg

First mass is heavier than first that is why mass second first will go downward and mass second will go upward.

Given that pulley and string is mass  less that is why both mass will have same acceleration.So lets take their acceleration is 'a'.

So now from force equation

m_1g-m_2g=(m_1+m_2)a

21.1 x 9.81 - 12.9 x 9.81 =(21.1+12.9) a

a=2.36\ m/s^2

Lets tension in string is T

m_1g-T=m_1a

T=m_1(g-a)

T=21.1(9.81-2.36) N

T=157.06 N

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The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

H = \frac{nI}{l}

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

n = \frac{Hl}{I}

The magnetic field is again given by,

H = \frac{B}{\mu_t}

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

n = \dfrac{\frac{B}{\mu_t}l}{I}

Replacing with our values we have that,

n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}

n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}

n = 27.5 \approx 28

Therefore the number of turn required is 28Truns

4 0
4 years ago
A square 1300-turn coil with the area of 0.2 m2 is situated in the xy-plane in a region where the magnetic field is (50k) μT. Wh
givi [52]

Answer:

e = 65 mA

Explanation:

given,

number of turn = 1300

Area = 0.2 m²

B = 50 x 10⁻⁶T

t = 0.2 s

coil rotate in x-z plane

θ = 90°  to θ = 0°

maximum induced emf

e = -\dfrac{BAn(cos \theta_f - cos\theta_i)}{\Delta t}

e = -\dfrac{1300 \times 0.2 \times 50\times 10^{-6}(cos 90^0- cos0^0)}{0.2}

e =0.065

e = 65 mA

7 0
4 years ago
A galloping pony speeds past you at 5 m/s. The frequency of the sound produced by the hooves on the dirt is 221 Hz. Assume the s
kow [346]
Given:
speed of passing pony 5 m/s
frequency of the sound produced: 221 Hz
speed of sound 342 m/s

Let us use the Doppler Shift Formula:
Where the <span>source is moving away from the observer at rest
</span>
f' = (v / v+vs) f
Where, vs<span> = Velocity of the Source,</span>
           v = Velocity of sound or light in medium,
           f = Real frequency,
           f' = Apparent frequency.

f'= [342 m/s / (342 m/s+5m/s)] * 221 Hz
f' = 0.9856 * 221Hz
f' = 217.8176 Hz or 218 Hz

The observed frequency <span>of the hooves after the pony has passed your position is 218 Hz.</span>
7 0
3 years ago
Read 2 more answers
4. What happens to the total resistance of a<br> parallel circuit when another resistor is added?
Dahasolnce [82]

Answer:add resistance in the parallel circuit with the total resistance

Explanation:

the total resistance in the parallel circuit is also add resistance.

3 0
2 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
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