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4 years ago

1. What are the two quantities involved when calculating MOMENTUM?

Physics

1 answer:

Snezhnost [94]4 years ago

8 0

Answer:

Mass and Velocity

Explanation:

Momentum = Mass X Velocity

or p=m' x v

since in physics p is quantity momentum and m stands for mass and v stands for velocity.

Hope this helps :)

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Yusef is doing an experiment to find out how the amount of sunlight affects the growth of a seedling what is the variable in yis

Mekhanik [1.2K]

Answer:

Independent Variable: Amount of Sunlight

Dependent Variable: Growth of Seedling

Constant: Things that don't change like kind of soil, type of pot, amount of water, etc.

Explanation:

I'm not sure which variable (independent, dependent, etc) so I'm going to do them all

7 0

3 years ago

Describe what we may observe from Earth regarding the moon’s revolution around the earth and its rotation

slava [35]

When we look at the moon from the Earth, we always see the same light spots, dark spots, and shapes. It never changes. There could be two possible reasons for this:

-- The moon is a flat disk with some markings on it, and one side of it always faces the Earth.

-- The moon is a round ball with some markings on it, and one side of it always faces the Earth.

Either way, since the same side always faces the Earth, the only way that can happen is if the moon's revolution around the Earth and rotation on its axis both take EXACTLY the same length of time.

Even if they were only one second different, then we would see the moon's whole surface over a long period of time. But we don't. So the moon's rotation and revolution must be EXACTLY locked to the same period of time.

4 0

3 years ago

Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One o

Zarrin [17]

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

qA = ½ q

qB = ½ q

qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

qA = 1/2 (q / 2) = ¼ q

qC = ¼ q

qB = ½ q

At this point all spheres have a charge,

qA = ¼ q

qb = ½ q

qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

Q = ½ q + ¼ q = ¾ q

When splitting the charge

qB = ½ ¾ q = 3/8 q

qC = ½ ¾ q = 3/8 q

qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

Q = qA + QB + QC

Q = ½ q + 3/8 q + ¼ q

Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

8 0

3 years ago

Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a)

mafiozo [28]

(a) The work done by the bird to raise the snake on the branch is equal to the product between the weight of the snake and the height of the branch above the ground:

$W_1 = mg h$

where $m=75 g=0.075 kg$ is the mass of the snake and $h=2.5 m$ is the height of the branch with respect to the ground. Substituting numbers into the equation, we find

$W_1 = (0.075 kg)(9.81 m/s^2)(2.5 m)=1.84 J$

(b) The work the bird did to raise its own centre of mass from the ground to the branch is equal to the product between the bird's weight and the height of the branch:

$W_2 = mgh$

where $m=350 g=0.35 kg$ is the mass of the bird. Substituting numbers into the equation, we find

$W_2 = (0.35 kg)(9.81 m/s^2)(2.5 m)=8.58 J$

6 0

3 years ago

Two electric charges qa = 1. 0 μc and qb = - 2. 0 μc are located 0. 50 m apart. how much work is needed to move the charges apar

melisa1 [442]

The total work done of 0.018 joules is needed to move the charges apart and double the distance between them.

We have two electric charges q(A) = 1μc and q(B) = -2μc kept at a distance 0.5 meter apart.

We have to calculate much work is needed to move the charges apart and double the distance between them.

<h3>What s the formula to calculate the Potential Energy of a system of two charges (say 'q' and 'Q') separated by a distance 'r' ?</h3>

The potential energy of the system of two charges separated by a distance is given by -

$U = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}$

In order to solve this question, it is important to remember the work - energy theorem which states -

"The change in the energy of the body is equal to work done on it"

Hence, using this work -energy theorem in the question given to us we get -

$U_{f} -U_{i} =W_{net}$

In our case -

$U_{f} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r}\\\\U_{i} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W=\frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r} - \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W = \frac{qQ}{4\pi\epsilon_{o}r} (\frac{1}{2} -1)\\\\W = 9\times 10^{9}\times \frac{1 \times 10^{-6} \times 2\times10^{-6} }{0.5} \times \frac{-1}{2}$

W = 0.018 joules

Hence, the total work done should be 0.018 joules.

To solve more question on potential energy, visit the link below -

brainly.com/question/15014856

#SPJ4

4 0

2 years ago