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2 years ago

What current, in units of milliAmperes, would produce a potential difference of 4.5

Physics

1 answer:

Bess [88]2 years ago

5 0

Answer;

= 56.96 mA

Explanation;

From the Ohm's law;

V = IR, where V is the potential difference in volts, I is the electric current in amperes and R is the resistance in ohms.

V = 4.5 volts, R = 79 Ω

Making I the subject of the formula we get;

I =V/R

= 4.5 /79

= 0.05696 A

But, 1 A = 1000 mA

Thus; 0.05696 × 1000

= 56.96 mA

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Ceres, Pluto, and Eris are all round in shape and classified as:_________ A) Leftover planetesimals that formed inside the frost

motikmotik

Answer

Ceres, Pluto, and Eris are classified as DWARF PLANET.

A) Leftover planetesimals inside the frost line are known as ASTEROIDS.

B) METEORITES are the pieces of Asteroids which are fallen on the earth's surface.

C) COMETS are the objects which are visible with long tails.

D) COMETS are also the leftover planetesimals that are occupied by the jovian planets and are formed in the solar system.

E) Meteor showers are associated with debris from COMETS

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2 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

3 years ago

Positive charge is distributed uniformly throughout a non-conducting sphere. The highest electric potential occurs: A. Far from

Ilya [14]

Answer:E

Explanation:

At the center

6 0

2 years ago

How much power is needed to lift the 200-N object to a height of 10 m in 4 s?

harkovskaia [24]

Answer:

500 watts

Explanation:

Recall that the definition of power is the amount of energy delivered per unit of time.

In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:

200 N x 10 m = 2000 Nm

then the power is the quotient of this potential energy divided the time it took to lift the object to that position:

Power = 2000 / 4 Nm/s = 500 Nm/s = 500 watts

6 0

2 years ago

The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf

Galina-37 [17]

Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| 
E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J 

After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:

E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m 

so 6.56×10^-7 m or better written 656 nm is in the visible spectrum

7 0

2 years ago