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mariarad [96]
2 years ago
9

A force of 100N is applied to move an object a horizontal distance of 20m to the right. The work done by this force on the objec

t is
Physics
1 answer:
horsena [70]2 years ago
5 0
WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
SO, 2000J
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A bus travels with an average velocity of 60 km per hour. how long does it take to cover a distance of 500km?
sergeinik [125]
V = 60km/h
S=500km
t = ?
v = S/t
t = S/v
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t ≈ 8,33 h
6 0
3 years ago
A sound wave travels in a straight line at a constant speed of 660 mph. What is
Fed [463]

Answer:

a = 0

Explanation:

The velocity of a sound wave in a straight line is 660 mph. The wave is moving with a constant velocity. It means that the change in velocity is equal to 0.

We know that acceleration of an object is equal to the change in velocity divided by time taken. But here the change in velocity is 0. As a result the acceleration of the sound wave is 0.

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At the surface of Venus the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is
yanalaym [24]

Answer:

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b) 86 atm

c) 645 m/s

Explanation:

See attachment for calculations on how i arrived at the answer

8 0
3 years ago
I need help in my physics class and show me how it’s done
Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)

Where | A | is the magnitude of the vector and \alpha is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle \alpha.

Vector A is in the 4th quadrant

So:

A_x = 6\\\\A_y = -6.5

So:

|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846

So:

Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)

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Option 4.

4 0
3 years ago
You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. what
pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector F_{net} = vector Fe + vector Fn,

⇒ |F_{net}| = \sqrt{}  F^{2} _{e } + F^{2}_n

⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ F_{net} = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

⇒ F_{net}= √2KqQ

Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.

To learn more about positive charges here

brainly.com/question/2903220

#SPJ4

8 0
2 years ago
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