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2 years ago

8

The diagram shows the Earth rotating on it's axis. The two star symbols show different locations on the surface... What would th

e sky look like to a person standing at point Y? *
Sunrise
Middle of the day (noon)
Sunset
Middle of the night (midnight)

2 answers:

rewona [7]2 years ago

8 0

It would look like Sunset

taurus [48]2 years ago

7 0

It would be mid-day (noon)

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Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}$

Since

$L_{su}= m_{su}*v_{su}*r_1$

then

$v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}$

$v_{su} = 19.44 m/s$

7 0

3 years ago

How do you find the change in potential energy

ratelena [41]

P.E = mgh

This is the formula for potential energy.

This is where m is mass, g is the acceleration due to gravity, and h is height.

All you have to do is multiply all these numbers together.

3 0

3 years ago

A cyclist going downhill is accelerating at 1. 2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is

mel-nik [20]

Answer:

$\boxed {\boxed {\sf v_i= 4 \ m/s}}$

Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

$v_f= v_i + at$

The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.

$v_f$ = 16 m/s
a= 1.2 m/s²
t= 10 s

Substitute the values into the formula.

$16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)$

Multiply.

$16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)$

$16 \ m/s = v_i + 12 \ m/s$

We are solving for the initial velocity, so we must isolate the variable $v_i$ . Subtract 12 meters per second from both sides of the equation.

$16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s$

$4 \ m/s = v_i$

The cyclist's initial velocity is <u>4 meters per second.</u>

6 0

2 years ago

I need help asap pls

tigry1 [53]

no.9 indeed all matters are made out of atoms which is smaller than protons and neutrons

6 0

3 years ago

A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30∘ above the horizontal. At th

lana66690 [7]

Answer:

a) In order to catch the ball at the level at which it is thrown in the direction of motion.

b)Speed of the receiver will be 7.52m/s

Explanation:

Calculating range,R= Vo^2Sin2theta/g

R= (20^2×Sin(2×30)/9.8 = 35.35m

Let receiver be(R-20) = 35.35-20= 15.35m

The horizontal component of the ball is:

Vox= Vocostheta= 20× cos30°

Vox= 17.32m/s

Time taken to coverR=35.35m with 17.32m/s will be:

t=R/Vox= 35.35/17.32

t= 2.04seconds

b)Speed required to cover 15.35m at 2.04seconds

Vxreciever= d/t = 15.35/2.04 = 7.52m/s

7 0

3 years ago

Read 2 more answers