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2 years ago

How does the sun’s gravitational attraction impact Earth’s motion?

Physics

1 answer:

Tasya [4]2 years ago

7 0

The Sun's gravitational pull keeps our planet orbiting the Sun in a nice nearly-circular orbit.

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Computer-enhanced x-rays are called __________ scans. A. TMS B. PET C. CT D. MRI

goldenfox [79]

Answer:

C. CT

Explanation:

It stands for Computed Tomography Scans.

3 0

2 years ago

Read 2 more answers

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

hram777 [196]

Answer:

$v_{2}=3.5 m/s$

Explanation:

Using the conservation of energy we have:

$\frac{1}{2}mv^{2}=mgh$

Let's solve it for v:

$v=\sqrt{2gh}$

So the speed at the lowest point is $v=7 m/s$

Now, using the conservation of momentum we have:

$m_{1}v_{1}=m_{2}v_{2}$

$v_{2}=\frac{1*7}{2}$

Therefore the speed of the block after the collision is $v_{2}=3.5 m/s$

I hope it helps you!

8 0

2 years ago

What creates more pressure?

Evgesh-ka [11]

It is C, gasses with less kinetic energy, i did this and i think i remember it was C

8 0

2 years ago

What is your acceleration while sitting in your chair. the latitude of corvallis is 44.4˚.?

marta [7]

You can start with the equations you know

a=v^2/r = (2pi*r/T)^2/r = 4pi^2r/T^2

Radius of earth (R) = 6378.1 km
Time in one day (T) = 86400 seconds
Latitude = 44.4 degrees

If you draw a circle and have the radius going out at a 44.4 degree angle above the center you can then find the r.

r=Rcos(44.4)
r=6378.1cos(44.4)
r= 4556.978198 km or 4556978 m

Now you can plug this value into the acceleration equation from above...

a= 1.8*10^8/7.47*10^9
a= .0241 m/s^2

8 0

2 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

2 years ago