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2 years ago

Look at the diagram of a solar eclipse.

Physics

1 answer:

lubasha [3.4K]2 years ago

3 0

Answer:

at location 4

Explanation:

it is because the solar eclipse happens when the moon is covering the sun

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Question 3. AP PHYSICS. Currently learning about torque so I assume you apply torque. I don't get it at all

8090 [49]

wow that is confusing

8 0

3 years ago

A 160.-kilogram space vehicle is traveling along a straight line at a constant speed of 800. Meters per second. The magnitude of

natulia [17]

Answer:

Zero

Explanation:

As force acting on the body is equal to the product of mass and acceleration.

Acceleration is equal to rate of change in velocity.

Here velocity is constant so acceleration is zero.

It means the net force acting on the vehicle is zero.

4 0

3 years ago

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An object is moving at 2.50 m/s [E]. At a time 3.00 seconds later the object is traveling at 1.50 m/s

babunello [35]

Given parameters:

First velocity = 2.50m/s

Time of travel = 3s

Second velocity = 1.50m/s

Unknown:

The displacement during the first interval = ?

Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.

Velocity = $\frac{Displacement}{Time taken}$

So;

Displacement = Velocity x Time taken

Now input the parameter for the first velocity and time of travel;

Displacement = 2.5 x 3 = 7.5m

The displacement id 7.5m

7 0

3 years ago

An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro

valkas [14]

Answer:

Final velocity of electron, $v=6.45\times 10^5\ m/s$

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, $u=4.52\times 10^5\ m/s$

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

$v^2=u^2+2as$ ........(1)

a is the acceleration, $a=\dfrac{F}{m}$

We know that electric force, F = qE

$a=\dfrac{qE}{m}$

Use above equation in equation (1) as:

$v^2=u^2+\dfrac{2qEs}{m}$

$v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m$

v = 647302.09 m/s

$v=6.45\times 10^5\ m/s$

So, the final velocity of the electron when it reaches point B is $6.45\times 10^5\ m/s$ . Hence, this is the required solution.

3 0

3 years ago

A graph titled Position versus time for with horizontal axis time (seconds) and vertical axis position (meters). The line runs i

Lelu [443]

Answer:

The first one is 3 m/s

The second one is 2 m/s

Explanation:

8 0

2 years ago

Read 2 more answers