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Jlenok [28]
2 years ago
7

A weight lifter lifts a 1.0 x 102 kg mass 1.5 m in 2.0 s. The power is?

Physics
1 answer:
Serga [27]2 years ago
6 0

Answer:

\boxed{p =  {7.4 \times 10}^{2} W} \to \: option \: c.

Explanation:

his \: power \: is \to \\ p =  \frac{mgh}{t}  \\ p =  \frac{1.0 { \times 10}^{2}  \times 1.5 \times 9.8}{2.0}   \\  p= 735 \\  \boxed{p =  {7.4 \times 10}^{2} }

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Which of the following happens to an object in uniform circular motion?
podryga [215]

Answer:

As an object moves in a circle, it is constantly changing its direction. ... Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed.

5 0
3 years ago
g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th
kati45 [8]

The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

3 0
2 years ago
If 5 mm of rain falls in a 100 m2 field, what volume of rain, in m3, fell in the field?
Nina [5.8K]

The volume of rain that fells in the field is simply given by the area of the field, which is

A=100 mm^2

multiplied by the height of rain that fell, which is

h=5.0 mm

Therefore, the volume is

V=hA=(5 mm)(100 mm^2)=500 mm^3

7 0
3 years ago
The lung capacity of the average adult is 5.5 liters at the surface and only .37 liters at a depth of 100 meters. The formula to
Alika [10]
I think you can just sub the values in? unless the qn is asking for smth else?

4 0
3 years ago
What was the result of the Mexican army's victory over the Texan garrison at the Alamo?
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3 0
2 years ago
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