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3 years ago

Explain the forces that keep a rock balanced on a tiny pedestal. Please help me with this!

1 answer:

8 0

Normal force for the rock because that makes an object stable at its position.
static friction because micro-welts hold its particle on its position so it doesn't change in position by a potential energy. Gravity makes it stay on the ground because its force attraction between an object and the earth.
Hope this helps <span />

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On a boat ride, the skipper gives you an extra-large life preserver filled with lead pellets. When he sees the skeptical look on

Lelu [443]

Answer:

True.

Explanation:

He is correct but what he does not tell is that you will be drown. Your life preserver will submerge and displace more water than those of your friends who float at the surface. Although buoyant force on you will be greater , the net downward force on you will still be greater. Hence you will be drown inside the water.

8 0

3 years ago

An emu moving with constant acceleration covers the distance between two points that are 92 m

nasty-shy [4]

Answer:

a) $V_{o}=14.30 m/s$

b) $a=-0.046 m/s^{2}$

Explanation:

The complete question is written below:

An emu moving with constant acceleration covers the distance between two points that are 92 m apart in 6.5s. Its speed as it passes the second point is 14 m/s. What are (a) its speed at the first point and (b) its acceleration?

Since we are talking about constant acceleration, we can use the following equations:

$d=x-x_{o}=(\frac{V_{o}-V}{2})t$ (1)

$V=V_{o}+at$ (2)

Where:

$d=92 m$ is the distance between the two points

$V_{o}$ is the velocity of the emu at the first point

$V=14 m/s$ is the velocity of the emu at the second point

$t=6.5 s$ is the time it takes to the emu to cover the distance $d$

$a$ is the emu's constant acceleration

Knowing this, let's begin with the answers:

<h2>a) Speed at the first point</h2>

In this situation wi will use equation (1):

$d=(\frac{V_{o}-V}{2})t$ (1)

Finding $V_{o}$ :

$V_{o}=\frac{2d}{t}-V$ (3)

$V_{o}=\frac{2(92 m)}{6.5 s}-14 m/s$ (4)

$V_{o}=14.30 m/s$ (5)

<h2>b) Emu's acceleration</h2>

Now we will substitute (5) in equation (2):

$14 m/s=14.30 m/s+a(6.5 s)$ (6)

Finding $a$ :

$a=-0.046 m/s^{2}$ (7) This means the emu is decreasing its speed at a constant rate.

8 0

3 years ago

Using the right-hand rule from your lessons, determine the directions of the electrical current and magnetic field of the electr

aliya0001 [1]

Answer:

Hello there use something that looks like this

Explanation:

This is an accurate representation of something you are working on!

As you can see the wire and the core are represented on the left and is showing how it can be represented on your right hand and how they are similar!