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3 years ago

A ball dropped from a height of 50 meters, Determine the speed of the ball after 3 seconds

Physics

1 answer:

Alex17521 [72]3 years ago

8 0

16.7 m/s

To determine the speed we use the formula: Distance ÷ Time so, 50 ÷ 3 = 16.66... ( rounded off to 1.d.p is 16.7).

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Please hurry Describe why electric currents can be dangerous

konstantin123 [22]

Answer:

Cardiac Arrest, burns, and nerve damage.

Explanation:

Basically, the main risk is cardiac arrest, caused by the electric current interfering with the normal operation of the heart muscle. Other possible damages are burns due to the electric energy vaporizing the water inside the cells, and nerve damage caused by excessive current through the nerves.

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2 years ago

PLEASE HELP! DUE DATE IS TODAY

g100num [7]

Answer:

Question 2

Part (a)

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Oxygen: type of compound = oxide

Part (b)

The iron reacts with water and oxygen to form rust.

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Question 3

This circuit is in parallel.

The current in a parallel circuit splits into different branches then combines again before it goes back into the supply.

We are told that A₁ = 0.8 A

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A₂ = 0.4 A

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Then combines again:

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2 years ago

An applied force of 50 N is used to accelerate an object, that weighs 73 N, to the right across a frictional surface. The object

Hunter-Best [27]

Answer:

5.38 m/s^2

Explanation:

NET force causing the object to accelerate = 50 -10 = 40 N

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F = ma

40 = 7.44 * a a = 5.38 m/s^2

6 0

1 year ago

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

a. The frequency with which the waves strike the hill is 242.61 Hz

b. The frequency of the reflected sound wave is 254.23 Hz

c. The beat frequency produced by the direct and reflected sound is

11.62 Hz

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

2 years ago

Which statement correctly describes the location, charge, and mass of the

Alenkinab [10]

The neutrons are inside the nucleus, have no charge, and have mass.

7 0

3 years ago