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1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms2. When an electric soldering iron is used in a 110 V circuit, the current flowing through the iron is
2 A. What is the resistance of the iron?
3. A current of 0.2 A flows through an electric bell having a resistance of 65 ohms. What must be
the applied voltage in the circuit?

Answers

Answer 1

Answer:

Answer:

(1) 0.04 ohms (2) 55 ohms (3) 13 volt

Explanation:

(1) The resistance of an electric device is 40,000 microhms.

We need to convert it into ohms.

$1\ \mu \Omega =10^(-6)\ \Omega$

To covert 40,000 microhms to ohms, multiply 40,000 and 10⁻⁶ as follows :

$40000 \ \mu \Omega =40000 * 10^(-6)\ \Omega\n\n=0.04\ \Omega$

(2) Voltage used, V = 110 V

Current, I = 2 A

We need to find the resistance of the iron. Using Ohms law to find it as follows :

V = IR, where R is resistance

$R=(V)/(I)\n\nR=(110)/(2)\n\nR=55\ \Omega$

(3) Current, I = 0.2 A

Resistance, R = 65 ohms

We need to find the applied voltage in the circuit. Using Ohms law to find it as follows :

V=IR

V = 0.2 × 65

V = 13 volt

Answer 2

Answer:

Answer:

1. 0.04 Ohms

2. 55 Ohms

3. 13 Volts

Explanation:

Penn Foster

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Why a switch is connected in phase wire and never is neutral wire?

Answers

If you had the wire connected to a neutral wire, you would never get a charge. You could receive a charge pulse from the phase wire.

Answer:

ifyou have a wre connect you should not have to connected

i think that is the answer

A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the lowest frequency associated with this portion of the spectrum?

Answers

Answer:

the lowest frequency is $7.5* 10^(14) Hz$

Explanation:

In the question it is given that wavelength(L) in the range of 200μm to 400μm.

let ν be frequency of wave v velocity = 3\times 10^8

velocity v= Lν

therefore ν= $(v)/(L)$

frequency ν be lopwest when L will be heighest

ν(lowest)= $(3* 10^8)/(400* 10^-9)$

ν= $7.5* 10^(14) Hz$

An object moving with uniform acceleration has a velocity of 10.5 cm/s in the positive x-direction when its x-coordinate is 2.72 cm. If its x-coordinate 2.30 s later is ?5.00 cm, what is its acceleration? The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity vi = 10.5 cm/s in the positive x-direction when the initial position is xi = 2.72 cm (t = 0). We are given that at t = 2.30 s, the final position is xf = ?5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. Xf-Xi=Vit-1/2at^2 Now we substitute the given values into this equation. (___cm)-(___cm)=(___cm/s)(__s)+1/2a(___s)

Answers

Answer:

Acceleration = 8.27 cm/s²

Explanation:

We are given;

initial velocity; v_i = 10.5 cm/s

Initial position; x_i = 2.72 cm

Time; t = 2.30 s

final position; x_f = 5.00 cm

To find the acceleration, we will make use of the formula;

x_f - x_i = (v_i * t) - (½at²)

Plugging in the relevant values, we have;

5 - 2.72 = (10.5 × 2.3) - (½ × a × 2.3²)

2.28 = 24.15 - 2.645a

24.15 - 2.28 = 2.645a

2.645a = 21.87

a = 21.87/2.645

a = 8.27 cm/s²

Using the kinematic equation, the acceleration of the object was calculated to be approximately8.27 cm/s² given its initial velocity, position, time, and final position.

We are given:

Initial velocity (vᵢ) = 10.5 cm/s

Initial position (xᵢ) = 2.72 cm

Time (t) = 2.30 seconds

Final position ( $x_f$ ) = 5.00 cm

We want to find the acceleration (a) of the object using the kinematic equation:

x₋ᵢ - xᵢ = (vᵢ * t) - (1/2) * a * t²

Now, let's substitute the given values:

5.00 cm - 2.72 cm = (10.5 cm/s * 2.30 s) - (1/2) * a * (2.30 s)²

Simplify the equation:

2.28 cm = 24.15 cm - (1/2) * a * 5.29 s²

Now, isolate 'a' by rearranging the equation:

-1.09 cm = (-1/2) * a * 5.29 s²

To remove the negative sign, multiply both sides by -1:

1.09 cm = (1/2) * a * 5.29 s²

Next, solve for 'a' by multiplying both sides by (2 / 5.29):

a ≈ (1.09 cm) / (2 / 5.29) s²

a ≈ 8.27 cm/s²

So, the acceleration of the object is approximately 8.27 cm/s².

For more such information on: acceleration

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A 2-C charge experiences a force of 40 N when put at a certain location inspace. The electric field at that location is a. 2 N/C.b. 20 N/C. c. 30 N/C. d.
40 N/C. e. 60 N/C.

Answers

Answer:

E = 20 N/C

Explanation:

Given that,

Charge, q = 2 C

Force experience, F = 40 N

We need to find the electric field at that location.

The electric field in terms of electric force is given by :

F = qE

Where

E is the electric field

$E=(F)/(q)\n\nE=(40\ N)/(2\ C)\nE=20\ N/C$

So, the electric field at that location is 20 N/C.

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .If the magnetic field isnt changed, what will be the orbital radius of the alpha particles? Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = (1)/(2)mv^(2)$

So, $v = \sqrt{(2E)/(m)}$ .... (1)

Now,

$r = (mv)/(Bq)$

Substitute the value of v from equation (1), we get

$r = (√(2mE))/(Bq)$

Let the radius of the alpha particle is r2.

For proton

So, $r_(1) = \frac{\sqrt{2m_(1)E}}{Bq_(1)}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_(2) = \frac{\sqrt{2m_(2)E}}{Bq_(2)}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$(r_(1))/(r_(2))=(q_(2))/(q_(1))* \sqrt{(m_(1))/(m_(2))}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$(r_(1))/(r_(2))=\frac{2q}}{q}}* \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

Answer:15.95 cm

Explanation:

Given

Energy=300 kev

radius of Proton=16 cm

mass of alpha particle $=6.64* 10^(-27) kg$

mass of proton $=1.67* 10^(-27) kg$

charge on alpha particle is twice of proton

radius of Proton is given by

$r=(mv)/(|q|B)$

and Kinetic energy $K=(P^2)/(2m)$

where P=momentum

$P=√(2Km)$

$r=(√(2km))/(qB)$ ---1

Radius for Alpha particle is

$r_(alpha)=\frac{\sqrt{2k\cdot m_(alpha)}}{2qB}$ -----2

Divide 1 & 2 we get

$(r)/(r_(alpha))=(√(m))/(q)* \frac{2q}{\sqrt{m_(alpha)}}$

$r_(alpha)=\sqrt{(6.64* 10^(-27))/(1.67* 10^(-27))}* 0.5$

$r_(alpha)=0.997* 16$

$r_(alpha)=15.95 cm$

Determine whether the following statements are true and give an explanation or counterexample.(A) If the acceleration of an object remains constant, its velocity is constant.
(B) If the acceleration of object moving along a line is always 0, then its velocity is constant.
(C) It is impossible for the instantaneous velocity at all times a(D) A moving object can have negative acceleration and increasing speed.

Answers

Answer:

Explanation:(A)if a body is accelerating then it's velocity can't be constant since an object is said to be accelerating if it is changing velocity (B)if the acceleration of an object moving along a line is 0 then it's velocity will be constant since there is no change in direction or speed(C)No.it is not possible for a moving body to have an instantaneous velocity at all times since instantaneous velocity is the velocity of a body at a certain instant of time..(D)Yes a moving object can have a negative acceleration and increasing speed,it can also have a positive acceleration with decreasing speed.

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