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Anarel [89]
3 years ago
13

From Doppler shifts of the spectral lines in the light coming from the east and west edges of the Sun, astronomers find that the

radial velocities of the two edges differ by about 4 km/s, meaning that the Sun’s rotation rate is 2 km/s. Find the approximate period of rotation of the Sun in days. The circumference of a sphere is given by 2πR, where R is the radius of the sphere
Physics
1 answer:
Gala2k [10]3 years ago
3 0

Answer:

T= 37 day

Explanation:

To solve this exercise we will use the definition of angular velocity as the angular distance, which for a full period is 2pi between time.

    w = T / t

The relationship between angular and linear velocity is

    v = w r

    w = v / r

We substitute everything in the first equation

    v / r = 2π / t

    t = 2π r / v

Let's reduce to the SI system

    V = 2 km / s (1000m / 1km) = 2 10³ m / s

    r= R = 6.96 10⁸ m

Let's calculate

    t = 2π 6.96 10⁸/2 10³

    t = 3.2 10⁶ s

    T = t = 3.2 10⁶ s ( 1h/3600s) (1 day/24 h)

    T= 37 day

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From the graph below, which segment has the fastest speed--A, B, C, D, or E?
zzz [600]

Answer:

D bcz the slope rose the fastest

Explanation:

I DON'T WANNA BE QUESTIONED, THIS IS BECAUSE y POSITION ROSE FASTER THAN x POSITION

8 0
3 years ago
One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large
svet-max [94.6K]

Given:

• Mass, m = 0.200 mg

,

• Speed, v = 3.00 x 10³ m/s

,

• Time, t = 6.00 x 10^⁻⁸ s.

Let's calculate the force exerted.

Using the inpulse-momentum theroerm, we have:

impulse = change in momemntum

Where:

Impulse = force x time

change in momentum = mass x velocity.

Thus, we have:

F\times6.00\times10^{-8}=(0.200\times10^{-6})\times(3.00\times10^3)

Let's solve for the force F:

\begin{gathered} F=\frac{(0.200\times10^{-6})(3.00\times10^3)}{6.00\times10^{-8}} \\  \\ F=10000N \end{gathered}

Therefore, the force exterted is 10000 N.

ANSWER:'

10000 N

4 0
1 year ago
What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving
lions [1.4K]

Complete question is;

Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.

What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?

Answer:

F = 501.5 N

Explanation:

We are given;

Mass of wooden crate; m = 75 kg

Angle of ramp; θ = 11°

Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;

F = mgsin θ + μmg cos θ

From online values, coefficient of friction between wooden surfaces is μ = 0.5

Thus;

F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)

F = 501.5 N

3 0
3 years ago
1. A truck of mass 3120 kg is accelerated from rest to a speed of 22.1 m/s. How much work was done on the truck to achieve that
Goryan [66]

Answer:

.5(3120kg)(22.1m/s)^2 = 7.62x10^5 J

Explanation:

8 0
3 years ago
The stoplight had just changed and a 2100 kg Cadillac had entered the intersection, heading north at 3.2 m/s , when it was struc
elena55 [62]

Explanation:

It is given that,

Mass of the Cadillac, m_1=2100\ kg

Speed of Cadillac, v_1=3.2\ m/s (towards north)

Mass of Volkswagen, m_2=1100\ kg

The cars stuck together and slid to a halt, leaving skid marks angled 35 degrees north of east.

According to the law of conservation of momentum,

The momentum along x axis, m_1v_1=(m_1+m_2)\ usin35............(1)

The momentum along y axis, m_2v_2=(m_1+m_2)\ ucos35...........(2)

From equation (1) and (2) it is clear that,

\dfrac{m_1v_1}{m_2v_2}=tan35

v_2=\dfrac{m_1v_1}{m_2\ tan35}

v_2=\dfrac{2100\times 3.2}{1100\times \ tan35}

v_2=8.72\ m/s

So, the Volkswagen going just before the impact is 8.72 m/s. Hence, this is the required solution.

6 0
3 years ago
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