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mezya [45]
2 years ago
14

A block with a weight of 620 N is pulled up at a constant speed on a very smooth ramp by a constant force. The angle of the ramp

with respect to the horizontal is θ = 23.5° and the length of the ramp is l = 14.1 m.
Calculate the work done by the force in pulling the block all the way to the top of the ramp. (Neglect friction.)

Physics
1 answer:
GuDViN [60]2 years ago
7 0

The work done by the force in pulling the block all the way to the top of the ramp is 3.486 kJ.

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

Work done is equal to product of force applied and distance moved.

Work = Force x Distance

Given is  a block with a weight of 620 N is pulled up at a constant speed on a very smooth ramp by a constant force. The angle of the ramp with respect to the horizontal is θ = 23.5° and the length of the ramp is l = 14.1 m.

From the Newton's law of motion,

ma =F-mg sinθ =0

So, the force F = mg sinθ

Plug the values, we get

F = 620N x sin 23.5°

F = 247.224 N

Work done by motor is  W= F x d

The force is equal to the weight F = mg

So, W = 247.224 x 14.1

W = 3.486 kJ

Thus, the work done by the force in pulling the block all the way to the top of the ramp is  3.486 kJ.

Learn more about Work done

brainly.com/question/13662169

#SPJ1

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Natalka [10]

Answer:  a. 17.7 KJ/Mol

b. T=210K

Explanation:

Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressure is 35 torr at – 111.95°c and 253 torr at – 83.6°c. using these data calculate.

the question isnt completely originally, but we could look at the likely derivation from the questions

(a) the standard enthalpy of vaporization

using the clausius clapeyron equation

In (PT1vap / PT2vap) = delta H (vap) / R ( (1/T1) - (1/T2) )

In (35Torr/253Torr) = delta H (vap) / 8.3145 ( (1/189.55) - (1/161.2) )

Therefore, Delta H (vap) = 17.7 KJ/Mol

b. Also the boiling point

What is the normal boiling point of arsine?

At the boiling point Pvap = atmospheric pressure = 1 atm=760 torr

substitution into the equation as stated in question 1

ln(760/253)=17700/8.314(1/189.55-1/T)

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5 0
3 years ago
MODERN PHYSICS
frosja888 [35]

Answer:

D. n=6 to n=2

Explanation:

Given;

energy of emitted photon, E = 3.02 electron volts

The energy levels of a Hydrogen atom is given as;  E = -E₀ /n²

where;

E₀ is the energy level of an electron in ground state =  -13.6 eV

n is the energy level

From the equation above make n, the subject of the formula;

n² = -E₀ / E

n² = 13.6 eV / 3.02 eV

n² = 4.5

n = √4.5

n = 2

When electron moves from higher energy level to a lower energy level it emits photons;

E = E_0(\frac{1}{n_1^2}-\frac{1}{n_2^2} )\\\\\frac{1}{n_1^2}-\frac{1}{n_2^2} = \frac{E}{E_o} \\\\\frac{1}{4} -\frac{1}{n_2^2} = \frac{3.02}{13.6} \\\\\frac{1}{4} -\frac{1}{n_2^2} =0.222\\\\\frac{1}{n_2^2} = 0.25 - 0.22\\\\\frac{1}{n_2^2} = 0.03\\\\n_2^2 = 33.33\\\\n_2 = \sqrt{33.33} = 6

For a photon to be emitted, electron must move from higher energy level to a lower energy level. The higher energy level is 6 while the lower energy level is 2

Therefore,  The electron energy-level transition is from n = 6 to n = 2

3 0
3 years ago
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tia_tia [17]

Answer:

4.54

Explanation:

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11X=50

X=4.54#

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Korvikt [17]
High density because it is obvious it will be hard to rip and tear.
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A 50-foot flagpole is at the entrance of a building that is 300 feet tall. If the length of the flagpole's shadow is 30 feet at
natka813 [3]
First you will want to sketch out both of the situations. It should be two sketches, one for the flagpole and one for the building.

To solve this, you will want to create a proportion.

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50/30= 300/x

Solve for x:
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7 0
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