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grigory [225]
3 years ago
10

Coal is lifted out of a mine a vertical distance of 50 m by an engine that supplies 500 W to a conveyer belt. How much coal per

minute can be brought to the surface? Ignore the effects of friction.
Physics
2 answers:
Brums [2.3K]3 years ago
7 0

Answer:

mass = 61.16 kg

Explanation:

given data

distance = 50 m

energy = 500 W = 500 × 60sec = 30000 J/min

to find out

mass

solution

we will apply here energy equation  

that is

energy = m×g×h    .............1

put here all value m mass here and g = 9.81 and h is distance

energy = m×g×h

30000 = m×9.81×50

mass = 61.16 kg

Nimfa-mama [501]3 years ago
3 0

Answer:

61 kg

Explanation:

Given data

distance (d): 50 m

power (P): 500 W = 500 J/s

mass (m): ?

time (t): 1 min = 60 s

The engine must do some work (w) to lift the coal. The work done in 1 minute is:

P = w / t

w = P . t = 500 J/s × 60 s = 3.0 × 10⁴ J

The work is equal to the force exerted (F) times the distance (d).

w = F × d

F = w / d = 3.0 × 10⁴ J / 50 m = 6.0 × 10² N

The force exerted is equal to the mass lifted (m) times the acceleration. Here, the acceleration to surpass is that of gravity (g = 9.8 m/s²).

F = m × g

m = F / g = 6.0 × 10² N / 9.8 m/s² = 61 kg

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Solve for the length of the inclined plane if the angle equals 19.45 degrees.
mel-nik [20]

The length of the inclined plane is approximately 12 ft

The situation forms a right angle triangle.

<h3>Right triangle</h3>

Right triangle have one of its angle as 90 degrees.

Therefore,

The length of the inclined plane is the hypotenuse of the triangle. The length of the inclined plane can be found using trigonometric ratios.

height = 4 ft

angle(∅) = 19.45°

sin 19.45 = 4 / h

h = 4 / 0.33298412235

h = 12.0125847796

h = 12 ft

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5 0
2 years ago
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
4 years ago
Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for the
WINSTONCH [101]

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

t_A=0.55s

t_B=0.45s

(a) The distance from the kicker to each of the 2 spectators is given by:

d_A=v \times t_A

where,

v= speed of sound

t_A=time taken for the sound waves to reach the ears

d_A=343\times 0.55=188.65m

(b)d_B=v \times t_B

where,

v= speed of sound

t_B=time taken for the sound waves to reach the ears

d_B=343\times 0.45=154.35m

(c)As the angle b/w slight lines  from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m

5 0
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Which of the following is most likely not a case of uniform circular motion
Ivan

Answer:

Motion of a racing car on a circular track

Explanation:

Uniform circular motion means the motion of the object is in a circle with a CONSTANT SPEED.

The racing car will accelerate during its motion. Hence, it is not a uniform circular motion.

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