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natulia [17]
3 years ago
12

a lemon with mass 0.3 kg falls out of a tree from a height of 1.8 m. How much mechanical energy does the lemon have just before

it hits the ground ?
Physics
2 answers:
Nadusha1986 [10]3 years ago
6 0
5.30 joules of mechanical energy. Hope I helped!
cestrela7 [59]3 years ago
6 0

Answer:

5.292 Joules

Explanation:

Hello

applying the law of the conservation of energy, and assuming that no energy is lost due to friction with the air, we can affirm that the energy that has just before the impact is the same energy that it has just before being released, that energy is due to height, mass and gravity and is defined by:

E_{g}=m*g*h\\\\now\\E_{p} =E_{g}=m*g*h\\E_{p}=m*g*h\\

Let

m=0.3kg\\g=9.8\frac{m}{s^{2} }\\ h=1.8\ m

pute the values into the equation

E_{p}=m*g*h\\E_{p}=0.3 kg*9.8\frac{m}{s^{2} }*1.8\m\\E_{p}=5.292\ Joules\\

Have a nice day.

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Answer:

False.

Explanation:

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Or, as expressed in mathematical terms:

\frac{a^3}{T^2}=constant, where <em>a</em> is the semi-major axis of the orbit (the distance from the center), and <em>T </em>is the orbital period of the satellite.

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A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to
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<span>The molecules in the thermometer’s liquid spread apart.  (B)</span>
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3 years ago
A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b
Natasha_Volkova [10]

Answer:

Part a)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

Outside the outer cylinder we will again use Guass law

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0
3 years ago
A train leaves the train station at noon and travels at a constant speed of vt = 50 mi/hr on a straight track. 2 hr later, a car
asambeis [7]

Answer:

350 miles

Explanation:

When the car starts 2 hours later, the train would have a head start of

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The speed of the car relative to the train is

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For the car to catch up with the train, it must cover the 100 miles difference at the rate of 20mi/hr. So the time it would need to cover this difference is

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mario62 [17]

Answer:

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Explanation:

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<em>where P: is the gas pressure, T: is the gas temperature and k: is a constant.</em>

Therefore, due to Gay-Lussac's Law, when the plunger is pushed down very rapidly, the pressure of the air increase, which leads to its temperature increase. That is why cotton flashes and burns.      

I hope it helps you!

3 0
3 years ago
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