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ryzh [129]
3 years ago
5

Suppose you find yourself in a spaceship in a uniform circular orbit around a star. Suppose also that that star is going to lose

half of its mass over some period of time (which would, of course, change the magnitude of the force of gravity keeping you in orbit). If you wanted to keep the same orbital speed in a uniform circular orbit after the star has only half of its current mass, how should you relocate your spaceship
Physics
1 answer:
neonofarm [45]3 years ago
3 0

Answer:

i actually have no idea...

Explanation:

i have no brain

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What is the natural pH of the ocean?
Katyanochek1 [597]

Answer: Basic

Explanation: water is "neutral" with a ph of 7, the ocean has like 8 so it is more basic on the scale

7 0
3 years ago
Read 2 more answers
Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much
shutvik [7]

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

6 0
2 years ago
A girl pulls on a 10-kg wagon with a constant horizontal force of 30 N. If there are no other horizontal forces, what is the wag
strojnjashka [21]
Force equals mass*distance
F = ma

Given m = 10 kg, F = 30 N

30 = 10a
30/10 = a
3 = a

The wagon's acceleration is 3 m/s^2
7 0
3 years ago
Please answer any of these thanks !
KIM [24]
1).  The equation is: (speed) = (frequency) x (wavelength)

Speed = (256 Hz) x (1.3 m) = 332.8 meters per second

 2).  If the instrument is played louder, the amplitude of the waves increases.
On the oscilloscope, they would appear larger from top to bottom, but the
horizontal size of each wave doesn't change.

If the instrument is played at a higher pitch, then the waves become shorter,
because 'pitch' is directly related to the frequency of the waves, and higher
pitch means higher frequency and more waves in any period of time.

If the instrument plays louder and at higher pitch, the waves on the scope
become taller and there are more of them across the screen.

3).  The equation is:  Frequency = (speed) / (wavelength)
(Notice that this is exactly the same as the equation up above in question #1,
only with each side of that one divided by 'wavelength'.)

Frequency = 300,000,000 meters per second / 1,500 meters = 200,000 per second.

That's ' 200 k Hz ' .

Note:
I didn't think anybody broadcasts at 200 kHz, so I looked up BBC Radio 4
on-line, and I was surprised.  They broadcast on several different frequencies,
and one of them is 198 kHz !
7 0
3 years ago
A projectile is fired with an initial velocity of 120.0 meters per second at an angle, θ above the horizontal. If the projectile
k0ka [10]

Answer:

θ = 62.72°

Explanation:

The projectile describes a parabolic path:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = x₀+ vx*t   Formula (1)

vx = v₀x

Where:  

x: horizontal position in meters (m)

x₀: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s

v₀x: Initial speed in x  in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)

vfy= v₀y -gt Formula (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 120 m/s  , at an angle  θ above the horizontal

v₀x= 55 m/s

x-y components of the initial  velocity ( v₀)

v₀x = v₀*cosθ Equation (1)

v₀y = v₀*sinθ   Equation (2)

Calculating of the angle θ

We replace data in the  Equation (1)

55 =  120*cosθ

cosθ = 55 / 120

\theta = cos^{-1}(  \frac{55}{120} )

θ = 62.72°

3 0
3 years ago
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