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4 years ago

Power is the time rate of change of work. a. True b. False

1 answer:

6 0

The statement about "<span>Power is the time rate of change of work" is true. The answer is letter A. Power is presented by the equation P = Wt where P is the power, W is work and t is time.</span>

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A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this

krok68 [10]

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

8 0

2 years ago

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

expeople1 [14]

Answer:

The value is $m \approx 310$

Explanation:

From the question we are told that

The focal length of the objective is $f_o = 1.0 \ cm$

The focal length of the eyepiece is $f_e = 2.0 \ cm$

The tube length is $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

$m = m_o * m_e$

Where $m_o$ is the objective magnification which is mathematically represented as

$m_o = \frac{L}{f_o }$

=> $m_o = \frac{25}{1 }$

=> $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

$m_e = \frac{L }{f_e }$

$m_e = \frac{25 }{ 2}$

$m_e = 12.5 \ cm$

$m = 25 * 12.5$

$m \approx 310$

6 0

3 years ago

Calculate the time needed for a 0.600 kg hammer to reach the surface of the Earth

USPshnik [31]

The time needed for the hammer to reach the surface of the Earth is 3.54 s.

<h3>Time of motion of the hammer</h3>

The time of motion is calculated as follows;

t = √(2h/g)

where;

h is height of fall
g is acceleration due to gravity

t = √(2 x 10 / 1.6)

t = 3.54 s

Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

8 0

2 years ago

Find the potential energy of a 2 kg ball 15 m in the air.

mart [117]

Answer:

294.3 Joules

Explanation:

2kg*9.81m/s^2*Δ15=294.3J

8 0

3 years ago

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

Softa [21]

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$

$t_2 = 5.08 sec$

from equation of motion we know that

$d_1 = vt_1 + \frac{1}{2} gt_1^2$

where d_1 is distance covered in time t1

so $d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2$ =

$d_1 = 110.78 m$

$d_2 = vt_2 + \frac{1}{2} gt_2^2$

where d_2 is distance covered in time t2

$d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2$

$d_2 = 213.31 m$

distance cover is = 213.31 - 110.78 = 102.53 m

3 0

3 years ago