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ss7ja [257]
3 years ago
11

An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindston

e is I = 2.0 kgm2 . The system starts from rest. A. Determine the angular acceleration of the shaft when this torque is applied. B. Determine the kinetic energy of the shaft after 8 seconds of operation. C. Determine the work done by the motor in this time. D. Determine the average power delivered by the motor over this time interval
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0

Answer:

Angular acceleration = 5 rad /s ^2

Kinetic energy = 0.391 J

Work done = 0.391 J

P =6.25 W

Explanation:

The torque is given as moment of inertia × angular acceleration

angular acceleration = torque/ moment of inertia

= 10/2= 5 rad/ s^2

The kinetic energy is = 1/2 Iw^2

w = angular acceleration/time

=5/8= 0.625 rad /s

1/2 × 2× 0.625^2

=0.391 J

The work done is equal to the kinetic energy of the motor at this time

W= 0.391 J

The average power is = torque × angular speed

= 10× 0.625

P = 6.25 W

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Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.
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Answer:

Explanation:

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7 0
3 years ago
Two pulses move in opposite directions on a string and are identical in shape except that one has positive displacements of the
Leni [432]

(c) When the two pulses completely overlap on the string forms a straight line.

A single disturbance that travels via a transmission medium is referred to as a pulse. This medium might be formed of stuff or a vacuum, and it might be endlessly large or finite in size.

Consider two pulses that are identical in shape and proceed in opposite directions along a string, with the exception that one has positive displacements of the string's elements while the other has negative displacements.

On the string, the two pulses blend together completely.

The pulses completely balance one another out in terms of removing string elements from equilibrium, yet the string still moves. Shortly after the string is once again shifted, the pulses will have passed each other.

The correct option is (c)

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6 0
1 year ago
Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s

5 0
3 years ago
If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?
SVETLANKA909090 [29]

B

Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is \sqrt{a^2 + b^2}

where a = 1.2 km north

and b = 1.6 km east

magnitude = 2 km

Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.

8 0
3 years ago
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