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WITCHER [35]
4 years ago
7

Mass of 0.6 kg height of 2.4 m how much gravitational potential energy

Physics
1 answer:
grin007 [14]4 years ago
8 0

Answer:

14 j

Explanation:

Protentional energy = PE = mgh

m = 0.6 kg

h = 2.4 m

g = 9.8 m/s²

PE= 0.6 kg * 9.8 m/s²* 2.4 m= 14 J

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A girl rides her scooter to school, a total distance of 4.5km. She has to slow down twice to cross busy streets, but overall the
Whitepunk [10]

Answer:

6.93 km/h

Explanation:

To calculate her average speed, we need the "speed" formula, which is:

average speed = distance / time

You plug in your numbers and it will give you the answer.

Speed = 4.5km/0.65hr

           = 6.923 km/h

8 0
3 years ago
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Do an Internet search to determine what minerals are extracted from the ground in order to manufacture the following products:
MrRa [10]

Answer:

Raw materials are most times gotten from the earth through various forms of extraction procedures.

A) Stainless steel utensils is made up of mainly Iron and other elements such as chromium , carbon etc.

B) Cat litter comprises of ceramic products which is made up of clay.

C) Tums brand antacid tablets comprises of calcium carbonate, magnesium hydroxide, aluminum hydroxide and sodium bicarbonate which could be extracted from the earth.

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E)Aluminum beverage cans are made up of aluminum extracted from the ground.

3 0
4 years ago
This involves convection currents.
Julli [10]
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6 0
3 years ago
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If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w
marshall27 [118]

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

5 0
4 years ago
a mass of 0.75 kg is attached to a spring and placed on a horizontal surface. the spring has a spring constant of 180 N/m, and t
Artemon [7]

Answer:

6.57 m/s

Explanation:

First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement

F=kx; F=180(.3) = 54 N

Next from Newton's second law find the acceleration of the mass.

Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²

Now use the kinematic equation for velocity (or speed)

v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.

v₀=0, since the mass is at rest before we release it

a=72 m/s² (from above)

x₀=0 as the start position already compressed

x₂=0.3m (this puts the spring back to it's natural length)

v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²

v₂=\sqrt{43.2)\\ = 6.57 m/s

5 0
4 years ago
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