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pentagon [3]
3 years ago
13

An object at rest has no net force acting on it. True False

Physics
2 answers:
enot [183]3 years ago
6 0

Answer:

True

Explanation:

Unless acted by another force a body moving at constant velocity or at rest remains at that state. So, yes an object at rest has no net force acting on it. Only when another force is acted on the object the object will move. This property is known as inertia. When the body is at rest it is known as inertia of rest and inertia motion when it is moving.

aivan3 [116]3 years ago
5 0
Yes that's true according to newton's first law 
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Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive
dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = \frac{-kQqi}{r (a+r)}

c.  F = \frac{-kQqi}{r^{2} }

Explanation:

a. x components:

dE = \frac{kdq}{(a+r-x^{2}) } \\

     = \frac{kQdx}{(a(a+r-x)^2}

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = \frac{-kQqi}{r (a+r)}

c. Given that r>>a, the expression becomes:

F = \frac{-kQqi}{r^{2} }

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

6 0
3 years ago
Two identical positive charges exert a repulsive force of 6.1 × 10−9 n when separated by a distance 3.6 × 10−10 m. calculate the
Natalka [10]
Im a jewish salesman wanting to know if you wanted to buy some jesus movies

3 0
2 years ago
An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the
Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

.

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

(n1)*(sin(theta1))=(n2)*(sin(theta2))

Then we will find the refraction angle relative to the vertical this way:

(n1/n2)*(sin(theta1))=sin(theta2)

(1/1.33)*(sin(45))=sin(theta2)

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

tan(32.12)=(X/30cm)

X=(tan(32.12))*(30cm)

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

tan(32.12)=(24cm/Y)

Y=(24cm/tan(32.12))

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm

5 0
3 years ago
Why is it important for engineers to design bumper cars safely
rosijanka [135]

Answer:

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Explanation:

3 0
3 years ago
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notsponge [240]

Answer:

Explanation:

Whacbdhnd

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3 years ago
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