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malfutka [58]
2 years ago
13

An object having a mass of 2.0 kilograms falls from a height of 15 meters. What is its kinetic energy when it hits the ground?

Physics
1 answer:
irakobra [83]2 years ago
4 0

The answer is 2.9 X 10 TO the 2 power.

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Describe each class of lever and explain to characteristics of each
Nataly [62]

-- Class I lever

The fulcrum is between the effort and the load.

The Mechanical Advantage can be anything, more or less than 1 .

Example:  a see-saw

-- Class II lever

The load is between the fulcrum and the effort.

The Mechanical Advantage is always greater than 1 .

Example:  a nut-cracker, a garlic press

-- Class III lever

The effort is between the fulcrum and the load.

The Mechanical Advantage is always less than 1 .

I can't think of an example right now.

8 0
2 years ago
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Randy wants to know whether a soil's porosity affects how easily seedlings grow in it.
telo118 [61]
Since the main objective of this experiment is to determine the effect of porosity on seedling growth that should be the only independent variable. In short, that is the only variable that should be different to ensure fair testing. 

The answer should be B:

he plants seedlings in soils with different levels of porosity and equal levels of permeability. 

Permeability is not what needs to be tested. If it changes, you may not be able to determine whether it was the porosity or permeability that cause changes.  
4 0
2 years ago
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If a magnet is broken into two pieces, what happens to the magnetic poles? One piece will have a north pole, while the other pie
Yuri [45]
Each piece will still have a North and South Pole
5 0
2 years ago
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
2 years ago
Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi
AleksandrR [38]
The horizontal component is calculated as:
Vhorizontal = V · cos(angle)

In your case Vhoriontal = 16 · cos(40) = 12.3 m/s

Answer: 12.3 m/s
7 0
2 years ago
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