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3 years ago

You arrive in science class 45 seconds after leaving math which is 90 meters to the south. What is your velocity?

1 answer:

6 0

Explanation: Well, If you arrive in my class 45 seconds after leaving math, which is 90 meters away. you traveled 2m/s ( I hope this helped <3)

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The acceleration due to gravity on earth will decrease as which of the following occurs. The mass of the object decreases. The d

pentagon [3]

Answer:

The distance of the object to the center of the earth increases.

Explanation:

The acceleration due to gravity on Earth is given by:

$g=\frac{GM}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the object from the Earth's centre

We notice that:

- g does not depend on the mass of the object

- g is inversely proportional to r

This means that if the distance of the object from the Earth's centre increases, g decreases. So, the correct option is

The distance of the object to the center of the earth increases.

4 0

3 years ago

The flow of cold dense air under the influence of gravity is called a _________ .

Yuki888 [10]

Answer: Katabatic Wind.

If you have any further questions, please comment. Have a great day. Thanks!
:)

6 0

3 years ago

Why would physics be used to study light emitted by a star?

VARVARA [1.3K]

Answer:

O D.

Explanation:

Physics has an aspect that deals with the study of energy

8 0

3 years ago

You wind a small paper tube uniformly with 153 turns of thin wire to form a solenoid. The tube\'s diameter is 5.11 mm and its le

lina2011 [118]

The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

$L = \frac{\mu N^2 A}{l}$

Here,

$\mu$ = Permeability at free space

N = Number of loops

A = Cross-sectional Area

l = Length

Replacing with our values we have,

$L = \frac{(4\pi *10^{-7})(153)^2(\pi (\frac{5.11*10^{-3}}{2})^2)}{2.47*10^{-2}}$

$L = 0.00002442H$

$L = 24.42\mu H$

Therefore the Inductance is $24.42\mu H$

3 0

3 years ago

A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th

12345 [234]

Answer:

Explanation:

<u>Given:</u>

Length of each side of the cube, $L=10\ cm$

The Elecric flux through one of the side of the cube is, $\phi =-1 kNm^2/C.$

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by $\epsilon_0$

Since Flux is a scalar quantity. It can added to get total flux through the surface.

$\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C$

So the the charge at the centre is calculated.

5 0

3 years ago