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3 years ago

How much energy is needed to melt 5 g of ice? The specific latent heat of melting for water is 334000 J/kg.

Physics

1 answer:

Katen [24]3 years ago

5 0

Answer:

The needed energy to melt of ice is 1670 J.

Explanation:

Given that,

Mass of ice = 5 g

Specific latent heat = 334000 J/kg

We need to calculate the energy

Using formula of energy

$Q=mL$

Where, m = mass

L = latent heat

Put the value into the formula

$Q=5\times10^{-3}\times334000$

$Q=1670\ J$

Hence, The needed energy to melt of ice is 1670 J.

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A circuit contains a 6.0-v battery, a 4.0-w resistor, a 0.60-µf capacitor, an ammeter, and a switch all in series. what will be

77julia77 [94]

<span>You are given a circuit that contains a 6.0-v battery, a 4.0 ohm resistor, a 0.60 micro farad capacitor, an ammeter, and a switch all in series. You are asked to find the current reading after the switch is closed. Apply ohms law where V = IR where V is the voltage, I is the current and R is the resistor.</span>

V = IR
I = V/R
I = 6 volts / 4 ohms
I = 1.5A

When the switch is closed, the cathode side plate begins to fill up with electrons when it was originally empty before the switch was closed. When it fills up the cathode side of the circuit, the current decreases. And when the capacitor cannot hold more electrons, the current will stop. The higher the capacitance, the higher is the capacity to store electrons.

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2 years ago

Read 2 more answers

Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They co

balu736 [363]

The combined momentum of the passengers is 5000 kgm/s.

<h3>Combined momentum of the passenger</h3>

The combined momentum of the passengers is calculated as follows;

P = mv1 + mv2

where;

m is mass of the passengers
v1 is velocity of the first passenger
v2 is velocity of the second passenger

P = m(v1 + v2)

P = 5000(-1 + 2)

P = 5000 kgm/s

Thus, the combined momentum of the passengers is 5000 kgm/s.

Learn more about momentum here: brainly.com/question/7538238

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1 year ago

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

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3 years ago

How does momentum conservation work for a rocket firing its engines to speed up in deep space?

Mashutka [201]

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3 years ago

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denis23 [38]

Answer:

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......

Hope this answer can help you.

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3 years ago