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earnstyle [38]
3 years ago
7

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s

haft by a cable connected to a winch. The shaft is inclined at 34.0° above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed.
Physics
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer:

11.714 kW

Explanation:

Here is the complete question

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 34.0∘ above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed. What power must the winch motor provide when the car is moving at constant speed?

Solution

Since the loaded ore car moves along the mine shaft at an angle of θ = 34° to the horizontal, if F is the force exerted on the cable, then the net force on the laoded ore car is F - mgsinθ = ma where  mgsinθ = component of the car's weight along the incline, m = mass of loaded ore car = 950 kg and a = acceleration

F = m(a + gsinθ)  

When the car is moving at constant speed, a = 0

So F = m(a + gsinθ) = F = 950(0 + 9.8sin34) = 5206.1 N

Since it continues at a constant speed of v = 2.25 m/s, the power of the winch motor is P = Fv = 5206.1 N × 2.25 m/s = 11713.7 W = 11.714 kW

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A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0
3 years ago
A wire is wrapped multiple times around a galvanized or aluminum nail and then each end of the wire connected to a battery for s
galina1969 [7]
Its A The paper clip is repelled away from the nail because an electromagnetic field magnetized to the nail
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3 years ago
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A lever is used to lift a heavy rock. The mechanical advantage of the lever is 4 and the lever applies a force of 800 N to the r
Lesechka [4]
 The force applied to the lever is 400 N, because the force applied by the lever (800 N) divided by the mechanical advantage of the lever (4) equals
400 N.

(800/4) = 200
7 0
3 years ago
two pendulums of lengths 100cm and 110.25cm start oscillating in phase. after how many oscillations will they again be in same p
goldfiish [28.3K]

Angular frequency of pendulum is given by

\omega = \sqrt{\frac{g}{l}}

for both pendulum we have

\omega_1 = \sqrt{\frac{9.81}{1.00}}

\omega_1 = 3.13 rad/s

For other pendulum

\omega_2 = \sqrt{\frac{9.81}{1.1025}}

\omega_2 = 2.98 rad/s

now we have relate angular frequency given as

[tex\omega_1 - \omega_2 = 3.13 - 2.98 = 0.15 rad/s[/tex]

now time taken to become in phase again is given as

t = \frac{2\pi}{\omega_1 - \omega_2}

t = \frac{2\pi}{0.15} = 41.88 s

now number of oscillations complete in above time

N = \frac{t}{\frac{2\pi}{\omega_1}}

N = \frac{41.88}{\frac{2\pi}{3.13}}

N = 21 oscillation


3 0
3 years ago
A car is traveling along a road, and its engine is turning over with an angular velocity of +188 rad/s. The driver steps on the
andreev551 [17]

Answer:

3751.80514 radians

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

\theta = Angle of rotation

t = Time taken

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=188\times 15.6+\dfrac{1}{2}\times 0\times 15.6^2\\\Rightarrow \theta=2932.8\ rad

The angular displacement would be 2932.8 rad

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=293\times 15.6+\dfrac{1}{2}\times 0\times 15.6^2\\\Rightarrow \theta=4570.8\ rad

The angular displacement would be 4570.8 rad

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{293-188}{15.6}\\\Rightarrow \alpha=6.73076\ rad/s^2

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\dfrac{293^2-188^2}{2\times 6.73076}\\\Rightarrow \theta=3751.80514\ rad

Actual value of the angular displacement is 3751.80514 radians

5 0
3 years ago
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