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Salsk061 [2.6K]
2 years ago
5

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.

70×107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 2.70×1011 m from its star with a period of 402 earth days. Once on the surface you find that the free-fall acceleration is 12.5 m/s2 .
A) What is the mass of the planet?
B) What is the mass of the star?
Physics
1 answer:
Arlecino [84]2 years ago
4 0

Answer:

sorry I don't know but you

Explanation:

can ask quora it's an app

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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

a.

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A²  ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

b.

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

c.

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J  = 120 J decrease

d.

The missing energy has turned into interned energy in the completely inelastic collision

3 0
3 years ago
How to calculate kinetic energy given mass and velocity
sashaice [31]

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

3 0
2 years ago
A rocket sled accelerates from 10 m/s to 40 m/s in 2 seconds. What is the average acceleration of the sled?
Dafna11 [192]

Answer:

15m/s²

Explanation:

Given parameters:

Initial velocity  = 10m/s

Final velocity  = 40m/s

Time taken  = 2s

Unknown:

Average acceleration  = ?

Solution:

Acceleration is the rate of change of velocity with time;

 Acceleration  = \frac{Final velocity   -  Initial velocity }{time}  

  Acceleration = \frac{40 - 10}{2}    = 15m/s²

6 0
3 years ago
Type the correct answer in the box. Spell the word correctly.
serious [3.7K]

Answer:

The Lambda-CDM model contains a cosmological constant, denoted by a lambda (λ), which is associated with dark energy and <u>cold dark matter</u>.

^Also works for Plato users.

3 0
3 years ago
Read 2 more answers
Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo
Elena-2011 [213]

a) 3.14 \cdot 10^{-4} s

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t] (1)

The generic equation that describes a wave is

y(t)=A sin (\frac{2\pi}{T} t) (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}

Therefore, the period is

T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when sin(...)=1, and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

\lambda=2L

where

\lambda is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

\lambda =2(5.00)=10.0 m

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]

And by comparing it with the general equation of a wave:

y(t)=A sin (\frac{2\pi}{T} t)

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

sin(\frac{2\pi}{T}t)=1

which means that

y(t)=A

And therefore in this case,

y=0.500 cm

So, this is the displacement.

6 0
3 years ago
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