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2 years ago

8

Mathew throws a ball straight up in the air. it rises for a period of time and then begins to drop. At which points in the balls

journey will be gravity be the greatest force acting on the ball

1 answer:

Allushta [10]2 years ago

3 0

Answer: The greatest force of gravity on the ball will occur at the point when the ball is near to hit the ground ... hope this helps have a good day :P

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Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

After the body has begun shivering, what explains why it would stop shivering?

Oksana_A [137]

The body shivers to produce energy and it uses the energy to keep it warm. The body would stop shivering when it has produced enough energy to keep it warm and the atmosphere around it has got warmer

3 0

3 years ago

Read 2 more answers

Why must the height of the meniscus in the graduated cylinder match the height of the water in the tub when measuring volume?

galben [10]

Answer:

See explanation

Explanation:

First, in order for you to understand, remember the basic concept of meniscus in graduated cylinder.

<em>"The meniscus is the curve seen at the top of a liquid in response to its container. The meniscus can be either concave or convex, depending on the surface tension of the liquid and its adhesion to the wall of the container".</em>

Now, according to this definition, and for water, the reading of the volume must be donde at the bottom of the curve of the meniscus. This is because the water gives a concave curve.

If you read it and matches the height of water, you are getting two results:

One, get an accurate value or volume, because it's been done at eye level.

The second fact is that when you do the reading this way, The total pressure is made equal to the atmospheric pressure by adjusting the height of the cylinder until the water level is equal.

8 0

3 years ago

An open pipe on an organ creates

PilotLPTM [1.2K]

Answer: 0.0163

Explanation: fn=n x v/2L

Fundamental frequency =f1

1 x (343/2L) = 10500

Rearrange the equation

L= v/(2xFn)

L= 343/(2x10500)

L=0.0163

8 0

2 years ago

2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas

Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

$ds=\dfrac{dQ}{dt}$

At constant pressure,

$dQ=C_{p}dt$

$\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}$

$\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})$

$\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}$

$0.693=ln\dfrac{T_{2}}{T_{1}}$

$ln2=ln\dfrac{T_{2}}{T_{1}}$

$T_{2}=2T_{1}$ ...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

$\Delta Q=C_{p}\Delta T$

Put the value into the formula

$6236=2.5\times8.3144(T_{2}-T_{1})$

$6236=20.786(T_{2}-T_{1})$

$T_{2}-T_{1}=\dfrac{6236}{20.786}$

$T_{2}-T_{1}=300$

Put the value of T₂

$2T_{1}-T_{1}=300$

$T_{1}=300\ K$

Put the value of T₁ in equation (I)

$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0

3 years ago