The blackbody curve for a star named Zeta is shown below. The most intense radiation for this star occurs in what spectral band?

A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th

Iteru [2.4K]

<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

$T^{2}=\frac{4\pi^{2}}{GM}R^{3}$ (1)

Where:

$G$ is the Gravitational Constant

$M=1.9(10)^{27}kg$ is the mass of planet X

$R$ is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

Solving:

$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>

3 0

3 years ago

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

mamaluj [8]

The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ = ¹/₂WL

T sinθ = ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

6 0

1 year ago

A convex lens can produce a real image but not a viral image<br> a. true<br> b. false

serious [3.7K]

The answer is true a convex lens can produce a real image but not a viral image

5 0

3 years ago

A giant armadilo moving northward with a constant acceleration covers the distance between two points 60m apart in 6 seconds. It

Naddika [18.5K]

Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. To determine acceleration, we need to know the initial velocity and the final velocity and the time elapsed. From the given values, we need t o calculate for the initial velocity. We use some kinematic equations. We do as follows:

x = v0t + at^2/2
60 = v0(6) + a(6)^2/2
60 = 6v0 + 18a (EQUATION 1)

vf = v0 + at
15 = v0 + a(6)
15 = v0 + 6a (EQUATION 2)

Solving for v0 and a,
v0 = 5 m/s
a = 1.7 m/s^2

8 0

3 years ago

Margaret, a researcher, desires to conduct a field experiment to determine the effects of a shopping mall's ambience on consumer

vichka [17]

Answer:

Extraneous

Explanation:

Extraneous variables are any variables that you are not intentionally studying in your experiment or test

6 0

3 years ago