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3 years ago

What is the difference between distance and position?

1 answer:

3 0

Distance is a scalar and measured only by units (meters, feet). Position is a vector and depends on where the origin is. It can be negative or positive and has direction.

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According to special relativity, in which frames of reference does light in a vacuum travel at less than 3.0 108 m/s?

ch4aika [34]

The correct option is D.
According to special relativity, in no frame of reference does light in a vacuum travel at less than the speed of light, the speed of light in a vacuum is the same for any inertial reference frame.This fact remain valid no matter the speed of a light source relative to another observer.

6 0

3 years ago

Read 2 more answers

The Global Positioning System (GPS) is a constellation of about 24 artificial satellites. The GPS satellites are uniformly distr

Liono4ka [1.6K]

Answer:

b) 3.72m/s²

c) 9.33*10^5

d) 9.33*10^5

e) 11.85 hrs

Explanation:

a) to confirm that gEarth is about 98 m/s².

Let's use the formula:

$gEarth= \frac{G*M}{R^2}$

$= \frac{6.67*10^-^1^1*5.972*10^2^4}{(6378*10^3)^2}$

= 9.78 m/s²

=> 9.8m/s²

b) Given:

$m = 6.417*10^2^3$

r = 2106 miles

$g_Mars = \frac{G*M}{R^2}$

$= \frac{6.67*10^-^1^1*6.417*10^2^3}{(2106*1.61*10^3)^2}$

=3.72 m/s²

c) we use:

$F = \frac{G*M*m}{R^2}$

$=\frac{6.67*10^-^1^1*5.972*10^2^4*1630*10^3}{((20000+6378)*10^3)^2}$

$= 9.33*10^5 N$

d) Let's take the force of gravitybon earth due to satellite as our answer in (c) because the Earth's gravitational force on a GPS satellite and the force of gravity on a GPS satellite on earth are equal and opposite (two mutual forces).

$F = 9.33*10^5 N$

e) In a circular motion,

Gravitional force = Centripetal force.

$\frac{GM*m}{R^2}=\frac{m*v^2}{R}$

$\frac{GM}{R}= v^2$

Solving for v, we have

$v= \sqrt{\frac{6*67*10^-^1^1*5.972*10^2^4}{(20000+6278)*10^3}}$

v = 3886m/s

Therefore,

v = 2πR/T

$3886 = \frac{2*pi*(20000+6378)*10^3}{T}$

Solving for T, we have:

T = 42650seconds

Convert T to hours

T = 42650/60*60

T = 11.86hrs

6 0

4 years ago

What do we call the certain electrons that are able to sometimes wander<br> away or be shared? *

Natalka [10]

Answer:

compond

Explanation:

5 0

3 years ago

5 examples of how intertia is used in real life

Mariulka [41]

Answer:

- your body moving forwards when sudden brakes/force is applied e.g. sudden braking in a car

- feeling a backwards force when something moves quickly from rest e.g a bus

- when you stir a cup of tea, it continues to swirl for a short period of time even though you have stopped

- when shaking a tree, it's leaves or fruit fall down

- satellites which move due to the inertia of motion

hope this helped :)

8 0

3 years ago

A string of length L, mass per unit length \mu, and tension T is vibrating at its fundamental frequency. What effect will the fo

viva [34]

The fundamental frequency on a vibrating string is given by:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension

$\mu$ is the mass per unit length of the string

Keeping this equation in mind, we can now answer the various parts of the question:

(a) The fundamental frequency will halve

In this case, the length of the string is doubled:

L' = 2L

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2(2L)}\sqrt{\frac{T}{\mu}}=\frac{1}{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{2}$

So, the fundamental frequency will halve.

(b) the fundamental frequency will decrease by a factor $\sqrt{2}$

In this case, the mass per unit length is doubled:

$\mu'=2\mu$

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}$

So, the fundamental frequency will decrease by a factor $\sqrt{2}$ .

(c) the fundamental frequency will increase by a factor $\sqrt{2}$

In this case, the tension is doubled:

$T'=2T$

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f$

So, the fundamental frequency will increase by a factor $\sqrt{2}$ .

8 0

3 years ago