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4 years ago

Which could cause a satellite to descend low enough that it burns up in the Earths atmosphere?

Physics

2 answers:

Bogdan [553]4 years ago

6 0

Answer:

B. Air resistance

Explanation:

If the distance at which the satellite orbits is not enough to avoid the atmospheric drag, it will be affected by the friction of the Earth's atmosphere and the drag will slow down the speed of the satellite. Therefore, It will result in the reduction in the altitude of a satellite's orbit and eventually burn.

Sati [7]4 years ago

5 0

The awenser is b air resistance

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Suppose a ray of light traveling in a material with an index of refraction n a reaches an interface with a material having an in

KATRIN_1 [288]

Answer: C and D

Explanation: One of the first rule for total internal reflection to occur is that the ray must move from a dense to a less dense medium, hence refractive index of medium a must be greater than that of b.

When a ray moves from a dense to a less dense medium, the refracted ray moves away from the normal thus increasing the size of the angle of refraction (total internal refraction occurs when the angle of refraction is 90° and the angle of incidence at this point is known as the critical angle), hence the angle of incidence must be greater than the critical angle.

These points verifies option C and D

5 0

3 years ago

A Ferris wheel turns at a constant 150.0 revolutions per hour. (a) Express this rate of rotation in units of radians per second.

RUDIKE [14]

Answer:

(a) 0.261 rad/s

(b) 1007.72 m

Explanation:

The angular velocity of the Ferris wheel is 150.0 revolutions per hour.

(a) To calculate the angular velocity of the wheel in units of radians per second, you take into account the following equivalence:

1 hour = 3600 seconds

1 revolution = 2π radians

You use the previous conversion factors:

$150.0\ \frac{rev}{h}*\frac{2\pi \ rad}{1\ rev}*\frac{1\ h}{3600\ s}=0.261\frac{rad}{s}$

In units of radians per seconds the wheel turns at 0.261 rad/s

(b) To find the arc length described by the wheel, you first calculate the angle described by the wheel in the time t, by using the following formula:

$\theta=\omega t$ (1)

ω: angular velocity = 0.261 rad/s

t: time = 4.95 min

You first convert the time to units of seconds

$4.95min*\frac{60s}{1min}=297s$

Next, you replace the values of the parameters in the equation (1):

$\theta=(0.261\frac{rad}{s})(297s)=77.51rad$

Next, you use the following formula for the arc length:

$s=r\theta$ (2)

r: radius of the wheel = 13.0 m

You replace the values of the parameters in the equation (2):

$s=(13.0m)(77.51rad)=1007.72m$

The arc length described by the wheel is 1007.72m

4 0

4 years ago

12 and 13 please ◀️⬆️⬇️↗️↩️⬅️↖️↔️

Black_prince [1.1K]

For 12 density often has unit of grams per cubic centimeter (g/cm3)

8 0

3 years ago

Using an example, explain how small atoms are.

Aliun [14]

Unlike a ball, an atom doesn't have a fixed radius. The radius of an atom can only be found by measuring the distance between the nuclei of two touching atoms, and then halving that distance... Does that answer your question?

4 0

3 years ago

A helix with 17 turns has height H and radius R. Charge is distributed on the helix so that the charge density increases like (i

marishachu [46]

The total charge on the helix is

Q = $\frac{221}{2}\sqrt{(\frac{H}{17})^2+(2\piR)^2$

Height of one turn = $\frac{H}{n}$

Horizontal distance in one turn = $2\pi R$

<h3>length of spring in one turn is</h3>

$l = \sqrt{h^2+(2\piR)^2}\\\\
l = \sqrt{(\frac{H}{n})^2+(2\piR)^2}
$

So, the total length of spring

$L = n*l\\\\
L = n * \sqrt{(\frac{H}{n})^2+(2\piR)^2}
$

Therefore, charge on spring

$Q = \frac{1}{2}*L*(13-0)\\\\
Q = \frac{1}{2}*n\sqrt{(\frac{H}{n})^2+(2\piR)^2}*13\\\\
Q = \frac{1}{2}*17*13\sqrt{(\frac{H}{17})^2+(2\piR)^2}\\\\
Q = \frac{221}{2}\sqrt{(\frac{H}{17})^2+(2\piR)^2\\\\
$

For more information on linear charge density, visit

brainly.com/question/15359786

8 0

2 years ago