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3 years ago

7

An electron drops from the n=6 to the n=4 level of an infinite square well that is 3.10-11 m wide. What is the wavelength of the

photon emitted as a result?
a.
0.148 nm

b.
0.025 nm

c.
0.327 nm

d.
1.205 nm

1 answer:

almond37 [142]3 years ago

4 0

The answer would be C !

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A person pulls a box across the floor with a rope. The rope makes an angle of 40 degrees tot he horizontal, and a total of 125 n

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Answer:

The angle formed of the rope with the surface = 40°

Force applied = 125Newtons

The displacement covered by the box =25metres

W= FDcos theta

[125×40×cos(40°) ] Joules

= [ (3125×0.76604444311)]Joules

= 2393.88888472 joules(ans)

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2 years ago

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2 years ago

An object is traveling on a circle with a radius of 6 feet. If in 80 seconds a central angle of 9/4 radians is swept out, then f

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Answer:

The angular speed of the object is 0.0281 rad/s

The linear speed of the object is 0.169 ft/s

Explanation:

Given;

radius of the circle, r = 6 ft

time of motion of the object around the circle, t = 80 s

central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad

The angular speed of the object is calculated as;

$\omega = \frac{\theta }{t} = \frac{2.25 \ rad}{80 \ s} = 0.0281 \ rad/s$

The linear speed of the object is calculated as;

v = ωr

v = 0.0281 rad/s x 6ft

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2 years ago

A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s

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Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

$F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\ 340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s$

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s

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2 years ago

A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the

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Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

F = p*V*g

F = 1000*(2*0.5*x*8*dx)*g

F = 78480*x*dx

- Now, the work done is given by:

W = F.s

- Where, s is the distance from top of hose to the differential volume:

s = (5 - x)

- We have the work as follows:

dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

W = 78400*(2.5*x^2 - (1/3)*x^3)

W = 78400*(2.5*3^2 - (1/3)*3^3)

W = 78400*13.5 = 1058400 J

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3 years ago