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3 years ago

6

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

ne of the equator. assuming the earth is a sphere with a radius of 6.38 106 m, determine the speed and centripetal acceleration of a person situated at each of the following.

1 answer:

Rama09 [41]3 years ago

7 0

Please elaborate more on your question so I can help you

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Is a spring stores 5 J of energy when its conpresses by 0.5 m what is the spring constant of the spring?

d1i1m1o1n [39]

Answer:

k = 40 N/m

Explanation:

A spring's energy is given:

$U = 0.5kx^2$

U is the energy in the spring, k is the spring constant and x is the spring displacement.

We are told that the spring stores 5J of energy, therefore, U = 5J. We are also told that the spring is compressed by 0.5m, so the spring x = 0.5m

$5 J = 0.5k(0.5m)^2\\5J = 0.5k(0.25m^2)\\5J = k*(0.125)m^2\\\\k = 5J/(0.125)m^2\\k = 40 N/m$

k = 40 N/m

Hope this helps!

3 0

4 years ago

HELP<br><br> These images represent waves. <br><br> Which wave has the highest frequency?

Ierofanga [76]

Wave C

because it is the closest or each wave is closer than the others

7 0

4 years ago

Read 2 more answers

Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what wil

drek231 [11]

Answer:The rate of ejection of photoelectrons will increase

Explanation:

If the frequency of incident monochromatic light is held constant and its intensity is increased, the rate of ejection of photoelectrons from the metal surface increases with increase in intensity of the monochromatic light. More current flows due to more ejection of photoelectrons.

4 0

3 years ago

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

lubasha [3.4K]

Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

$M = 4.54 \times 10^{23} kg$

Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

$\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}$

$\rho = 12580.7 kg/m^3$

8 0

3 years ago

In the following figure, if AB ǁ CD, then find the measure of PCD and CPD.

mina [271]

Answer:

$CPD = 80$

$PCD = 44$

Explanation:

Given

$AB || CD$

$BAD = 56$

$CPA = 100$

See attachment

Required

Determine PCD and CPD

First, we need to calculate CPD

Since DPA is a straight line and CPA = 100;

We have that:

$CPA + CPD = 180$ --- angle on a straight theorem

Substitute 100 for CPA

$100 + CPD = 180$

Subtract 100 from both sides

$100-100 + CPD = 180-100$

$CPD = 80$

Next, we calculate PCD

We have that:

$DAB= ADC = 56$ --alternate angle

In triangle PCD

$PCD + CPD + PDC = 180$ --- angles in a triangle

Where

$PDC = ADC = 56$

So, we have:

$PCD +80 + 56 = 180$

$PCD +136 = 180$

Subtract 136 from both sides

$PCD = 180 - 136$

$PCD = 44$

6 0

3 years ago