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2 years ago

What information does a free-body diagram show?

Physics

1 answer:

melamori03 [73]2 years ago

8 0

Answer:

Explanation:

In physics and engineering, a free body diagram (force diagram, or FBD) is a graphical illustration used to visualize the applied forces, moments, and resulting reactions on a body in a given condition.

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A and B, move toward one another. Object A has twice the mass and half the speed of object B. Which of the following describes t

BARSIC [14]

Answer:D

Explanation:

Given

mass of A is twice the mass of B half the velocity of B

Suppose $F_a$ and $F_b$ be the average force exerted on A and B respectively

and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.

Thus $F_a=-F_b$ and option d is correct

4 0

3 years ago

A 15.0 cm object is 12.0 cm from a convex mirror that has a focal length of -6.0 cm. What is the height of the image produced by

alisha [4.7K]

<h2>Answer: 5 cm</h2>

In convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object .

The Mirror equation is:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ (1)

Where:

$f=-6cm$ is the focal distance

$u=12cm$ is the distance between the object and the mirror

$v$ is the distance between the image and the mirror

We already know the values of $f$ and $u$ , let's find $v$ from (1):

$v=\frac{u.f}{u-f}$ (2)

$v=\frac{(12cm)(-6cm)}{12cm-(-6cm)}$

$v=-4cm$ (3)

On the other hand, the magnification $m$ of the image is given by the following equations:

$m=-\frac{v}{u}$ (4)

$m=\frac{h_{i}}{h_{o}}$ (5)

Where:

$h_{i}$ is the image height

$h_{o}=15cm$ is the object height

Now, if we want to find the image height, we firstlu have to find $m$ from (4), substitute it on (5) and find $h_{i}$ :

Substituting (3) in (4):

$m=-\frac{-4cm}{12cm}$

$m=\frac{1}{3}$ (6)

Substituting (6) in (5):

$\frac{1}{3}=\frac{h_{i}}{15cm}$

$h_{i}=\frac{15cm}{3}$

Finally we obtain the value of the height of the image produced by the mirror:

$h_{i}=5cm$

6 0

3 years ago

Read 2 more answers

1. Suppose you have a metal bar. Its mass is 57.9g and its volume is 3cm'. What is its<br> density?

mariarad [96]

¡Hellow!

How we know, for calculate density, we can use the formule:

$\boxed{\textbf{p = m / v} }$

<u>Where:</u>

$\sqrt{$ p = Density = ?

$\sqrt{$ m = Mass = 57,9 g

$\sqrt{$ v = Volume = 3 cm³

Let's replace according we information:

p = 57,9 g / 3 cm³

p = 19,3 g/cm³

The density of the metal bar is <u>19,3 g/cm³.</u>

<u></u>

¿Good luck?

3 0

3 years ago

Name the four major forces in the universe that act over long distance as I greater than the nucleus of an atom?

GalinKa [24]

Forces in the universe that act over long distance, meaning the distance is greater than the diameter of the nucleus of the atom are:

1. Electrostatic force or Coulomb force: Fc=(k*Q₁*Q₂)/r²,

2. Gravitational force: Fg=(G*m₁*m₂)/r²,

3. Magnetic force: Fm=qvB,

4. London dispersion force, also known as one of the van der Waals forces.

4 0

3 years ago

Why is it possible for a gas to take the shape of its container?

Sedaia [141]

I believe the answer is A

8 0

2 years ago

Read 2 more answers