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julia-pushkina [17]
3 years ago
15

A diagram of a closed circuit with power source on the left labeled 30 V, a resistor on the top labeled 2 Ohms, a resistor on th

e right labeled 3 Ohms and a resistor on the bottom labeled R Subscript 3 baseline.
If the current in this circuit is 3 A, what must be the value of R3?
Physics
2 answers:
sesenic [268]3 years ago
6 0

Answer:

5

Explanation:

larisa86 [58]3 years ago
3 0

Answer:

5

Explanation:

i dont have one

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The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
Which of these components is present in this circuit schematic?
Svetlanka [38]
Its resistor :}...............

4 0
3 years ago
Read 2 more answers
Jennifer, who has a mass of 50.0 kg, is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to av
Assoli18 [71]

Answer:

Average force = 3.5 kN

Explanation:

Given:

Mass of Jennifer (m) = 50 kg

Initial velocity = 35 m/s

Time taken to stop body = 0.5 s

Find:

Average force

Computation:

v = u + at

0 = 35 + a(0.5)

Acceleration (a) =  - 70 m/s² = 70 m/s²

Average force = ma

Average force = (50(70)

Average force = 3500 N

Average force = 3.5 kN

6 0
3 years ago
A small plastic bead has been charged to -15nC.
marishachu [46]

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

(C) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.  

8 0
3 years ago
A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 780 N on him. The
stich3 [128]

Answer:

f = 556.4 N

Explanation:

Mass of the losing player with its all equipment is given as

M = 86 kg

Net force applied on him by another player is given as

F = 780 N

also we know that acceleration of the losing player is given as

a = 2.6 m/s^2

now by Newton's 2nd law we will have

F - f = ma

780 - f = 86(2.6)

780 - f = 223.6

f = 556.4 N

7 0
3 years ago
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