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horrorfan [7]
3 years ago
12

Determine the magnitude of the current flowing through a 4.7 kilo ohms resistor if the voltage across it is (a) 1mV (b) 10 V (c)

4e^-t
Physics
1 answer:
mariarad [96]3 years ago
5 0

Answer:

213 nA

2.13 mA

851e^-t μA

Explanation:

We have a pretty straightforward question here.

Ohms Law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in the circuit. It is mathematically written as

V = IR, since we need I, we can write that

I = V/R

a) at V = 1 mV

I = (1 * 10^-3) / 4.7 * 10^3

I = 2.13 * 10^-7 A or 213 nA

b) at V = 10 V

I = 10 / 4.7 * 10^3

I = 0.00213 A or 2.13 mA

c) at V = 4e^-t

I = 4e^-t / 4.7 * 10^3

I = 0.000851e^-t A or 851e^-t μA

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Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
Which subatomic particle has a positive charge options:
poizon [28]
Proton is the answer ur looking for. have a wonderful day :) !!
3 0
3 years ago
Read 2 more answers
Consider the following 3 cases. (1) A block initially at rest on a floor is given a quick push by a hand. (2) The hand does not
jek_recluse [69]

Answer:

(1) The block isn't moving, despite the hand's force on it.

  • The net force on the block is zero.
  • The acceleration of the block is also zero.

(2) The block is gradually slowing down as it slides across the floor.

  • The acceleration of the block points in the opposite direction of the block's movement (i.e., the opposite direction of the block's velocity.)
  • The net force on the block points in the opposite direction of the block's movement. (Same direction as the block's acceleration.)

(3) Once again, the block is not moving.

  • The net force on the block is zero.
  • The acceleration of the block is also zero.

Explanation:

By Newton's Second Law, the net force on an object is in the same direction as its acceleration. Since this question said a lot about the object's motion, the direction of the object's acceleration might be easier to find than its net force.

<h3>(1)</h3>

The acceleration of an object is the rate of change of its velocity over time.

In this situation, the velocity of the object is zero, which is itself a constant. As a result, the rate of change of the object's velocity over time would be zero. Hence, the acceleration of the object would also be equal to zero. The zero vector doesn't have a specific direction.

By Newton's Second Law, the net force on the object is proportional to its acceleration. As a result, the net force on the object in this case would also be equal to zero. What about the hand's force on the block? The friction and normal force from the ground balances that force while the block hasn't yet started to move.

<h3>(2)</h3>

The object is slowing down over time. In other words, its velocity is decreasing over time. When a scalar value is decreasing, its rate of change would be negative. However, since velocity and acceleration are vectors, the acceleration of the object would be in the opposite direction of its velocity.

The net force on an object is in the same direction as its acceleration. As a result, the net force on this block in this case would also be in the opposite direction of the block's velocity.

<h3>(3)</h3>

Similar to the first situation, since the velocity of the block is zero (a constant,) its acceleration would be equal to zero. Since the net force on an object is proportional to its acceleration, the net force on this block would also be equal to zero.

7 0
3 years ago
Consider a air-filled parallel-plate capacitor with plates of length 8 cm, width 5.52 cm, spaced a distance 1.99 cm apart. Now i
prohojiy [21]

Answer:

The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

Explanation:

Given that,

Length of plates = 8 cm

Width = 5.52 cm

Distance = 1.99 cm

Dielectric constant = 2.6

Length = 4.4 cm

Potential = 0.8 V

We need to calculate the initial capacitance

Using formula of capacitance

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times8\times5.52\times10^{-4}}{1.99\times10^{-2}}

C=1.96\times10^{-12}

We need to calculate the final capacitance

Using formula of capacitance

C'=\dfrac{\epsilon_{0}A_{1}}{d}+\dfrac{k\epsilon_{0}A_{2}}{d}

Put the value into the formula

C'=(\dfrac{8.85\times10^{-12}}{1.99\times10^{-2}})((4.4\times5.52)+(3.6\times5.52)2.6)\times10^{-4}

C'=3.37\times10^{-12}

We need to calculate the  ratio of the new potential energy to the potential energy before the insertion of the dielectric

Using formula of energy

\dfrac{E}{E'}=\dfrac{\dfrac{1}{2}CV^2}{\dfrac{1}{2}C'V^2}

Put the value into the formula

\dfrac{E}{E'}=\dfrac{1.96\times10^{-12}}{3.37\times10^{-12}}

\dfrac{E}{E'}=0.58

Hence, The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

4 0
3 years ago
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