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2 years ago

Objects in space are moving at a constant velocity in a straight line.

Physics

1 answer:

tigry1 [53]2 years ago

8 0

Answer:

Explanation:

An object in motion will stay in motion unless acted on by a net positive or negative force.

For answer A. If the object were to be in an orbit, it would inevitably accelerate due to it being acted on by the gravitational force from the object it is orbiting. At different points in the orbit, the object will move at different speeds and continuously transfer between kinetic and potential energy.

For answer B. The object would would not stop their motion. In order for the object to lose energy, it would have to transfer it through friction or through its interaction with a gravitational field.

For answer D. No energy is "required" to maintain constant motion unless the object is willingly fighting against a resistive force like friction or a graviational well.

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What is refractive index

BARSIC [14]

Answer:

Explanation:

In optics, the refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as where c is the speed of light in vacuum and v is the phase velocity of light in the medium or for short the ratio of the velocity of light in a vacuum to its velocity in a specified medium.

4 0

3 years ago

Mike's car, which weighs 1,000 kg, is out of gas. Mike is trying to push the car to a gas station, and he makes the car go 0.05

Maru [420]

Answer:

The answer to your question is: F = 50 N

Explanation:

Data

mass = 1000 kg

acceleration = 0.05m/s2

F = ?

Formula

F = m x a

Substitution

F = 1000 kg x 0.05 m/s2 = 50 kgm/s2 = 50 N

Mike is applying a force of 50 N to the car.

3 0

3 years ago

An emu moving with constant acceleration covers the distance between two points that are 92 m

nasty-shy [4]

Answer:

a) $V_{o}=14.30 m/s$

b) $a=-0.046 m/s^{2}$

Explanation:

The complete question is written below:

An emu moving with constant acceleration covers the distance between two points that are 92 m apart in 6.5s. Its speed as it passes the second point is 14 m/s. What are (a) its speed at the first point and (b) its acceleration?

Since we are talking about constant acceleration, we can use the following equations:

$d=x-x_{o}=(\frac{V_{o}-V}{2})t$ (1)

$V=V_{o}+at$ (2)

Where:

$d=92 m$ is the distance between the two points

$V_{o}$ is the velocity of the emu at the first point

$V=14 m/s$ is the velocity of the emu at the second point

$t=6.5 s$ is the time it takes to the emu to cover the distance $d$

$a$ is the emu's constant acceleration

Knowing this, let's begin with the answers:

<h2>a) Speed at the first point</h2>

In this situation wi will use equation (1):

$d=(\frac{V_{o}-V}{2})t$ (1)

Finding $V_{o}$ :

$V_{o}=\frac{2d}{t}-V$ (3)

$V_{o}=\frac{2(92 m)}{6.5 s}-14 m/s$ (4)

$V_{o}=14.30 m/s$ (5)

<h2>b) Emu's acceleration</h2>

Now we will substitute (5) in equation (2):

$14 m/s=14.30 m/s+a(6.5 s)$ (6)

Finding $a$ :

$a=-0.046 m/s^{2}$ (7) This means the emu is decreasing its speed at a constant rate.

8 0

3 years ago

Accelerating charges radiate electromagnetic waves. calculate the wavelength of radiation produced by a proton in a cyclotron wi

levacccp [35]

B = magnetic field in the cyclotron = 0.400 T

q = magnitude of charge on a proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷ kg

f = frequency of revolution of proton in the cyclotron = ?

v = speed of electromagnetic waves = 3 x 10⁸ m/s

λ = wavelength of electromagnetic wave = ?

Frequency of revolution of proton in the cyclotron is given as

f = qB/(2πm)

inserting the values

f = (1.6 x 10⁻¹⁹)(0.400)/(2 (3.14) (1.67 x 10⁻²⁷))

f = 6.1 x 10⁶ Hz

wavelength of electromagnetic wave is given as

λ = v/f

λ = (3 x 10⁸)/(6.1 x 10⁶)

λ = 49.2 m

7 0

3 years ago

Read 2 more answers

2. You will need a magnifying glass and a small piece of scrap paper for this demonstration. a. Go outside on a sunny day and pu

andrew-mc [135]

The focal length of a magnifying glass is the distance between the focal point and optical centre of the magnifying glass.

<h3>Focal length</h3>

The focal length, f is the distance from a lens or mirror to the focal point, F.

This is the distance from a lens or mirror at which parallel light rays will meet for a converging lens or mirror or appear to diverge from for a diverging lens or mirror.

A magnifying glass is a converging lens which produces a enlarged, erect and virtual image when an object is placed between the focal point and optical centre.

A magnifying glass will bring to focus at a point sun rays which can cause the paper to catch fire if it is held in place for long.

This point at which the most concentrated ray of light is shining on the paper, is the focal point for that magnifying glass.

Therefore, the focal length of a magnifying glass is the distance between the focal point and optical centre of the magnifying glass.

Learn more about about focal length at: brainly.com/question/25779311

4 0

2 years ago