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KIM [24]
2 years ago
11

I was walking with Ruby in a garden from rest in a straight line with uniform acceleration so that we covered 0.25 m in the fift

h second of our motion. Calculate our acceleration. ​
Physics
1 answer:
Serhud [2]2 years ago
7 0

Answer:

0.02m/s^2

Explanation:

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Amotor supplied by 240V requires 12A to lift a 2000 lb at a rate of 25 ft/min, me power input to the motor is-
tensa zangetsu [6.8K]

Answer:

Power input, P = 2880 watts

Explanation:

It is given that,

Voltage of the motor, V = 240 V

Current required, I = 12 A

Weight lifted, W = 2000 lb

It is lifting at a speed of 25 ft/min. We need to find the power input to the motor. The product of current and voltage is called power input of the motor.

P=I\times V

P=12\ A\times 240\ V

P = 2880 watts

So, the power input of the motor is 2880 watts. Hence, this is the required solution.

4 0
3 years ago
An object releasing energy is evidence of which type of chemical change?
Ray Of Light [21]
Exothermic is the answer to your question
8 0
2 years ago
A high powered rifle can shoot a bullet at a speed of 1500 mi/hr. On the moon, with almost no atmosphere and an acceleration due
Amiraneli [1.4K]

Answer:

Explanation:

On the Moon :----

1500 x 1.6 = 2400 m /s is initial velocity of bullet .

g = 1.6 m /s²

v = u - gt

0 = 2400 - 1.6 t

t = 1500 s

This is time of ascent

Time of decent will also be the same

Total time of flight = 2 x 1500 = 3000 s

On the Earth : ---

v = u - a₁ t

0 = u - a₁ x 18

u = 18a₁

v² = u² - 2 x a₁ x 2743.2

0 = (18a₁ )² - 2 x a₁ x 2743.2

a₁ = 16.93

For downward return

s = ut + 1/2 a₂ x t²

2743.2 = 0 + .5 x a₂ x 31²

a₂ = 5.7 m /s²

If d be the deceleration produced by air

g + d = 16.93 ( during upward journey )

g - d = 5.7

g = (16.93 + 5.7) / 2  

= 11.315 m / s

d = 5.6 m /s²

So air is creating a deceleration of 5.6 m /s².

6 0
3 years ago
Miguel's parents drove him to a track meet that was 305 km away from the school. It took them 3 hours, which included a 30 minut
OverLord2011 [107]
A., 101.7 km/h is the correct answer for this question
6 0
3 years ago
A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2
il63 [147K]

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

f_o = \frac{v - v_o}{v}f

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz

The frequency heard by the observer is 993.52 Hz.

4 0
3 years ago
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