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3 years ago

Drawing a ray diagram for an object far from convex lens requires how many rays?

Physics

1 answer:

Crank3 years ago

5 0

Answer

Explanation:

four rays can also explain the ray diagram if the object is at infinity

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A 28 kg student stand on the surface of the Earth. What is the magnitude of the

Contact [7]

Answer:

F = 274.68[N]

Explanation:

The gravitational force is equal to the weight of a body, or this case that of a person. Weight can be calculated by means of the product of mass by gravitational acceleration. In this way we have the following equation:

$F=m*g$

where:

F = force or weight [N]

m = mass = 28 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

$F=28*9.81\\F=274.68[N]$

4 0

3 years ago

What is the acceleration of a 30.0-kg go kart if the thrust of its engine is 300.0 N?

VikaD [51]

Answer : $a=10\ m/s^2$

Explanation :

It is given that,

Mass of the engine, m = 30 kg

Thrust is equivalent to the force acting perpendicularly and it is F = 300 N

According to Newton's second law of motion :

$F = m\times a$

a is the acceleration of the engine.

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

$a=\dfrac{300\ N}{30\ Kg}$

$a=10\ m/s^2$

So, the acceleration of the engine is $10\ m/s^2$ .

Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago

____________ are used to calculate the distance a continent has moved in a year. a. Space satellites c. Laser beams b. Mirrors d

Law Incorporation [45]

<span> Space satellites, laser beams, mirrors</span> are used to calculate the distance a continent has moved in a year.

Therefore, your correct answer would be "all of the above".

4 0

3 years ago

What makes the planets gravity?

anyanavicka [17]

The presence of mass makes gravity. Doesn't matter whether it's a planet, a black hole, a puppy, or a speck of dust.

3 0

2 years ago