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exis [7]
1 year ago
14

Assume that the ammeter in the figure below is removed and the current that flows through the 4.0Ω path, I3, is unknown. Determi

ne all the currents in the circuit.

Physics
1 answer:
qaws [65]1 year ago
8 0

The circuit is in parallel connection

Equivalent resistance = 1/Req = 1/R1 + 1/R2 + 1/R3

From the information given,

R1 = 5

R2 = 2

R3 = 4

1/Req = 1/5 + 1/2 + 1/4 = (4 + 10 + 5)/20 = 19/20

Req = 20/19 = 1.053 ohms

I = V/R

Given that V = 12,

Current flow through circuit = 12/1.053 = 11.4 A

I1 + I2 + I3 = 11.4

I1 = 12/5 = 2.4 A

I2 = 12/2 = 6 A

I3 = 12/4 = 3A

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A person standing at the edge of a seaside cliff kicks a stone horizontally over the edge with a speed of 18 m/s. The cliff is 5
Fed [463]

Answer:

it would take 3.26 seg for the stone to fall to the water

Explanation:

If we ignore air friction then:

h=h₀ + v₀*t -1/2*g*t²

where

h= coordinates of the stone in the y axis ( height of the stone relative to the surface of the water )

h₀ = initial coordinates of the stone ( height of the cliff relative to the surface of the water = 52 m )

v₀ = initial <u>vertical </u>velocity = 0 ( since the ball is kicked horizontally , has only initial horizontal velocity , and has 0 vertical velocity )

t = time to reach a height h

g = gravity = 9.8 m/s²

since v₀ =0

h= h₀ - 1/2*g*t²

h₀ - h =  1/2*g*t²

t= √[2(h₀ - h)/g]

when the stone hits the ground h=0 ( height=0) , then replacing values

t=√[2(h₀ - h)/g]=√[2(52 m- 0 m )/(9.8m/s²)] = 3.26 seg

t= 3.26 seg

it would take 3.26 seg for the stone to fall to the water

7 0
3 years ago
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
frutty [35]

Answer:

Velocity=1.1m/s

Amplitude=0.35m

Explanation:

Given:

time 't' = 2.9s

wavelength 'λ'= 5.5m

distance 'd'=0.7m

The time period 't' is the time b/w two successive waves. Therefore, the time it takes from the boat to travel  from its highest point to its lowest is a half period.

So, T = 2 x 2.9 => 5.8 s

As we know that frequency is the reciprocal of time period, we have

f= 1/T = 1/5.8 =>0.2 Hz

In order to find how fast are the waves traveling, the velocity is given by

Velocity = f λ

V= 0.2 x 5.5 =>1.1m/s

The distance between the boat's highest point to its lowest point is double the amplitude.

Therefore , we can write

Amplitude 'A'= d/2 =>0.7/2 =>0.35m

8 0
3 years ago
A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made
anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2}  } }      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

B=\frac{4\pi\times10^{-7}\times  0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2}  } }

B = 6.99 x 10⁻⁶ T

3 0
3 years ago
State the formula for calculating power
Alex
Where power<span> P is in watts, voltage V is in volts and current I is in amperes (DC).</span>Power Formula<span> 2 – Mechanical </span>power equation<span>: </span>Power<span> P = E ⁄ t where </span>power<span> P is in watts, </span>Power<span> P = work / time (W ⁄ t). Energy E is in joules, and time t is in seconds.</span>
7 0
3 years ago
A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When
malfutka [58]

In order to determine the acceleration of the block, use the following formula:

F=ma

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

F=kx

Then, you have:

ma=kx

by solving for a, you obtain:

a=\frac{kx}{m}

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}

Hence, the acceleration of the block is 10 m/s^2

8 0
1 year ago
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