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ExtremeBDS [4]
3 years ago
9

What is the gravitational potential energy of a 0.1 kg apple sitting on top of a 12.5 m flagpole? Note: GPE= mgh. m= mass, g= ac

celeration due to gravity (9.8 m/s^2),
h= objects height above the ground.
Physics
1 answer:
Pachacha [2.7K]3 years ago
6 0
<span>GPE= mgh = 0.1 kg * 9.8 m/s^2  * 12.5m = 12.25 J</span>
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4. How much mass is required to exert a force of 25 Newtons, accelerating at 5 m/s?
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F = ma
F/a = m
(25 N) / (5 m/s2) = m
4 kg = m
8 0
3 years ago
Neptune's largest satellite is named WHAT and revolves in a WHAT orbit.
Rom4ik [11]
Neptune's largest satellite is Triton, and it revolves in a retrograde orbit, meaning to opposite of Neptune's totation<span />
4 0
3 years ago
What are carbon-12, carbon-13, and carbon-14?
denis23 [38]

Answer:

They are all Carbon isotopes.  Carbon 13 has 6 protons and 7 neutrons

Explanation:

Hope this was helpful! :)

4 0
3 years ago
A 0.350 kg block at -27.5°C is added to 0.217 kg of water at 25.0°C. They come to equilibrium at 16.4°C. What is the specific he
Gekata [30.6K]

The specific heat capacity of the block is 508J/kg^{\circ}C

Explanation:

As the block is placed into the water, heat energy is transferred from the water (which is at higher temperature) to the block (which is at lower temperature), until the block and the water are in thermal equilibrium (= same temperature).

Therefore, we can write:

Q_{water}=Q_{block}

Where

Q_{water}=m_w C_w (T_w-T_{eq}) is the heat energy released by the water, where

m_w = 0.217 kg is the mass of the water

C_w = 4186 J/kg^{\circ}C is the water heat specific capacity

T_w = 25.0^{\circ} is the initial temperature of the water

T_{eq}=16.4^{\circ} is the temperature at equilibrium

Substituting,

Q_{water}=(0.217)(4186)(25.0-16.4)=7812 J

Now we can write the heat energy absorbed by the block as

Q_{block}=m_b C_b(T_{eq}-T_b)

where

m_b=0.350 kg is the mass of the block

C_b is the specific heat capacity of the block

T_b = -27.5^{\circ} is the initial temperature of the block

And solving for C_b,

C_b=\frac{Q_{block}}{m_b(T_{eq}-T_b)}=\frac{7812}{(0.350)(16.4-(-27.5))}=508J/kg^{\circ}C

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

6 0
3 years ago
You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.80 mA when a sin
Anika [276]

Answer:

2.36 x 10^6 Hz

Explanation:

We know ;,

Iamp = 1.8mA

L = 0.45mH

VL = 12V

VL = IwL

VL = I x 2π x f x L

f = VL / (I x 2π x L)

f = 12 V/ (1.8mA x 2π x 0.45mH)

f = 12/(1.8x10^-3 x 2π x 0.45 x 10^-3)

f = 2.36 x 10^6 Hz

5 0
4 years ago
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