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596277

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Posted: 13 years ago

#481

Light from the sun reaches Earth in 8.3 min. The speed of light is 3.00 * 10^8m/s. How far is Earth from the sun?

Okay so I did this whole thing and I got 1.494*10^-11m, and do I have to convert this to Km? Does it matter?

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596277

Suspended

Posted: 13 years ago

#482

I need help with this distance question. Refer to this graph to find the distance the moving object travels between a. T = 0s and t = 5s.

b. t = 5s and t = 10s

c. t = 10s and t = 15s.

D = 0s and t = 25s.

Okay so for the first one I got 150 using the formula d = vt. I multiplied 30 the velocity by 5 seconds the time. Would this be right?

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akhl

@akhl

Dazzler

Posted: 13 years ago

#483

Originally posted by: -Ayeshaa-
Hey, I need help with this physics equation. I think I did the first half correct I just need help solving for g. If you could show it on paint or something that would be great.

http://imageshack.us/f/717/solveforg.jpg/

I am using V to mean square root.

T = 2 pi V(L/g)
Squaring both sides,
T^2 = 4 pi^2 L/g
g = 4 pi^2 L/T^2

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akhl

@akhl

Dazzler

Posted: 13 years ago

#484

Originally posted by: -Ayeshaa-
Light from the sun reaches Earth in 8.3 min. The speed of light is 3.00 * 10^8m/s. How far is Earth from the sun?

Okay so I did this whole thing and I got 1.494*10^-11m, and do I have to convert this to Km? Does it matter?

You have put -11 as exponent, it should be 11
1.494 * 10^11 m

No need to convert to m unless question asks you to do so. Keeping in m is fine enough.

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akhl

@akhl

Dazzler

Posted: 13 years ago

#485

Originally posted by: -Ayeshaa-

I need help with this distance question. Refer to this graph to find the distance the moving object travels between a. T = 0s and t = 5s.

b. t = 5s and t = 10s

c. t = 10s and t = 15s.

D = 0s and t = 25s.

Okay so for the first one I got 150 using the formula d = vt. I multiplied 30 the velocity by 5 seconds the time. Would this be right?

Easier way would be to find area below the graph in each case.
a. The graph between t=0s and t=5s is a triangle of base 5-0 = 5 s and height 30 m/s
Distance = area of the triangle = 1/2 * base * height = 1/2 * 5 * 30 = 75 m

b. Between 5s and 10s, the area below the graph is a rectangle of base 10-5 = 5 s and height 30 m/s
Distance = area of the rectangle = base * height = 5 * 30 = 150 m

c. Between 10s and 15s, the area below the graph is a trapezium of parallel sides 30 m/s and 20 m/s and distance between the parallel sides 5 s.
Distance = area of the trapezium
= 1/2 * (sum of parallel sides) * distance between parallel sides
= 1/2 * (30+20) * 5
= 125 m

d. Distance
= area(from 0s to 5s) + area(from 5s to 10s) + area(from 10s to 15s) + area(from 15s to 20s) + area(from 20s to 25s)
= 75 m + 150 m + 125 m + 5*20 + 1/2*5*20
= 500 m

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Sonia

@Miggi

Inactive

Posted: 13 years ago

#486

Thank you for help. :) i have one more question.

A device called pitot tube is used to measure the airspeed of a plane. A simplified model of the tube is a manometer with one side connceted to a tube facing directly in the wind. On that particular side of the tube( the static chamber), the air is motionless. The other side is connected to a tube into which the air moves at the speed of the plane. This air flows right past small holes, which are used to communicate the decrease in the pressure on the othe side of the manometer. If the manometer uses mercury and the levels differ by 25cm, what is the airplane speed in km/hr? The density of the air at the plane's altitude is taken to be 0.90kg/m^3. How does the speed you get compare to what is posted on the internet of commercial airliner's cruising speeds?

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596277

Suspended

Posted: 13 years ago

#487

Thanks so much Akhl :)

Jenny has a mass of 35.6 kg and her skateboard has a mass of 1.3 kg. What is the momentum of Jenny and her skateboard together if they are going 9.50m/s?

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akhl

@akhl

Dazzler

Posted: 13 years ago

#488

Originally posted by: -Ayeshaa-
Thanks so much Akhl :)

Jenny has a mass of 35.6 kg and her skateboard has a mass of 1.3 kg. What is the momentum of Jenny and her skateboard together if they are going 9.50m/s?

Momentum = m v
= (35.6 + 1.3) * 9.50
= 351 kg.m/s

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596277

Suspended

Posted: 13 years ago

#489

^ thank you so much:)

Two bowling balls each have a mass of 6.8kg. They are placed next to one another with their centres 21.8cm apart. What gravitational force do they exert on each other?

^ For that do I use the formula Fg = Gm1m2/d^2

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596277

Suspended

Posted: 13 years ago

#490

T = 2 pi V(L/g)
Squaring both sides,
T^2 = 4 pi^2 L/g
g = 4 pi^2 L/T^2

For this, for the squaring part, you squared the L and G and pi? You squared them altogether? So didn't have to divide by pi to get rid of it?

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