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2 years ago

1- an example on a body that it is in motion to an observer and at rest to another.

Physics

1 answer:

vesna_86 [32]2 years ago

6 0

Answer:

Explanation:

On a train you are sitting so not in motion but to someone outside the train you are in motion with the train

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Compare and contrast the way molecules behave in liquid to the way they behave in solids and gases.

Vitek1552 [10]

In liquids, molecules particles are close together but move in random directions.

In solids, they are closely packed together and they vibrate a little.

In gases, they love around really quickly and are farther apart (even more than liquids).

3 0

2 years ago

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The inductive reactance of the circuit is exactly twice the resistance: XL=2R. Adjust the phasor that represents the voltage acr

SVETLANKA909090 [29]

Answer:

∅= $63.43^{0}$

Explanation:

z=impedance

$x_{l}$ =2R

R=R

The resultant of the resistances in the circuit is called impedance

$x_{l}$ is inductive reactance of the circuit

R is the resistance of the resistor

z= $\sqrt{xl^{2}+R^2 }$

z= $\sqrt{2R^{2}+R^2 }$

Z= $\sqrt{5R^2}$

Z=R $\sqrt{5}$ ohms

tan∅=2R/R

tan∅=2

∅=Tan^-1(2)

∅= $63.43^{0}$

phase angle is ∅= $63.43^{0}$

3 0

2 years ago

You are in a planet where the acceleration due to gravity is known to be 3.28 m/s^2. You drop a ball and record that the ball ta

notsponge [240]

B) 7.87 m/s

The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;
$a = \frac{final \: velocity - initial \: velocity}{time \: taken}$

Given:

• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2

We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.

$3.28 = \frac{final \: velocity - 0}{2.40}$
$3.28 \times 2.40 = final \: velocity$
$final \: velocity = 7.872 \frac{m}{ {s} }$

5 0

3 years ago

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

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3 years ago

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Mama L [17]

Answer:

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3 years ago

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