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3 years ago

How much energy is needed to melt 150 g of ice at 0°C to water? (1)(Lf =3.34˟ 10⁵ J/Kg)

Physics

1 answer:

NARA [144]3 years ago

5 0

Answer:

5.01×10⁴ J.

Explanation:

Applying,

q = Cm....................... Equation 1

Where q = amount of heat needed to melt the ice, m = mass of the ice, C = specific latent heat of ice.

From the question,

Given: m = 150 g = (150/1000) kg = 0.15 kg, C = 3.34×10⁵ J/kg

Substitute these values into equation 1

q = (0.15×3.34×10⁵)

q = 0.501×10⁵ J

q = 5.01×10⁴ J.

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What is the density of a iphone with a mass of 200g and a volume of 40cm3

Leviafan [203]

Answer: 5 gm/cc

Explanation:

200 gm/40 cc
= 5 gm/cc

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3 years ago

Students perform an experiment in which they drop two eggs with equal mass from a balcony. in the first trial, the egg hits the

german

The impulse was greater in the first experiment because the egg broke.

<h3>What is impulse?</h3>

The term impulse is defined a the product of the force and time. We know that the impulse is high when a large force acts for a short time.

From the experiment if the students, we can conclude that the impulse was greater in the first experiment because the egg broke.

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2 years ago

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of

Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

$E = \frac{mg}{q}$

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

$q = \frac{mg}{E}$

Replacing,

$q = \frac{(3.37*10^{-9})(9.8)}{11000}$

$q = 3.002*10^{-12}C$

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

$q = ne$

Here,

n = Number of electrons

e = Charge of each electron

$n = \frac{q}{e}$

Replacing,

$n = \frac{3.002*10^{-12}}{1.6*10^{-19}}$

$n = 2.44*10^7$

Therefore the number of electrons that reside on the drop is $2.44*10^7$

5 0

4 years ago

Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope

Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

= 26,754 (0.816)

= 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c. What is the work done by the gravitational force on the sled?

By using the formula:

Gravity work = mgdsinθ

= 97.5 kg * 9.8 m/s² * 28 m * sin 60

= 23,170 J

23,170 J the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done = -1337.3 + 21,835 + 23,170

= 43,670 J

The net work done on the sled, in joules is 43,670 J.

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4 0

2 years ago

Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the wo

klemol [59]

Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy ( $E_{k}$ ) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

$E_{k}$ = h(f − f₀)

$E_{k}$ = hf - hf₀

E is the energy of the absorbed photons: E = hf

ϕ is the work function of the surface: ϕ = hf₀

$E_{k}$ = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy $E_{k}$ = 4.16×10⁻¹⁷ J

Speed of light c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js

E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

E = 53.8356 x 10⁻¹⁶ J

from $E_{k}$ = E - ϕ ;

ϕ = E - $E_{k}$

ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J

4 0

3 years ago