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dezoksy [38]
3 years ago
5

How to do this question?​

Physics
2 answers:
marissa [1.9K]3 years ago
7 0

Answer:

i3 =11.014A

i5 = 3.15A

Explanation:

Here according to k'chofs first law

i1 =i2 + i3

i3 = i4 + i5

For determine the i1 you have to consider the resultant resistor of the system

4 , 1 and 3 resistors are in pararel

Then, Resultant is

1/4 + 1/1 + 1/3 = 1/ R

R = 12/19

For get total we have to add another remaining 3 resistor because of serious

Then Resultant is = 12/19 + 3

= 69/19

Then using V = IR

40 =i3* 69/19

i3 = 11.014 A

Other 3 resistors are parrarel because of this voltage of those resistors are same.

Then i inversely propotional to its resistor

Then ,

i5 * 2 = (i3-i5)*4/5

i 5 = 3.15 A

Diano4ka-milaya [45]3 years ago
6 0

Answer:

I₁ = 11.2 A

I₂ = 1.6 A

I₃ = 9.6 A

I₄ = 6.4 A

I₅ = 3.2 A

Explanation:

Find the equivalent resistances.

Resistors 4 and 5 are in parallel:

R₄₅ = 1 / (1/R₄ + 1/R₅)

R₄₅ = 1 / (1/1 + 1/2)

R₄₅ = 2/3

Resistor 2 is in parallel with that:

R₂₄₅ = 1 / (1/R₂ + 1/R₄₅)

R₂₄₅ = 1 / (1/4 + 3/2)

R₂₄₅ = 4/7

Resistor 1 is in series with that:

R = R₁ + R₂₄₅

R = 3 + 4/7

R = 3 4/7

Now use Ohm's law to find I₁.

V = I₁ R

40 V = I₁ (3 4/7)

I₁ = 11.2 A

Find the voltage drop across R₁:

V₁ = I₁ R₁

V₁ = (11.2 A) (3 Ω)

V₁ = 33.6 V

So the voltage after is 40 V − 33.6 V = 6.4 V.

Find I₂:

6.4 V = I₂ (4 Ω)

I₂ = 1.6 A

Find I₃:

6.4 V = I₃ (2/3 Ω)

I₃ = 9.6 A

Find I₄:

6.4 V = I₄ (1 Ω)

I₄ = 6.4 A

Find I₅:

6.4 V = I₅ (2 Ω)

I₅ = 3.2 A

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An astronaut stands by the rim of a crater on the Moon, where the acceleration of gravity is 1.62 m/s2 and there is no air. To d
Nesterboy [21]

Answer:

12.1 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement = 120 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

Equation of motion

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -1.67\times 120-0^2\\\Rightarrow u=\sqrt{2\times 1.67\times 120}\\\Rightarrow u=20.02\ m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20.2}{-1.67}\\\Rightarrow t=12.1\ s

Time taken by the rock to hit the bottom of the crater is 12.1 seconds

5 0
3 years ago
A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder
Stella [2.4K]

Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

Angular frequency = 150 rpm

Stroke = 2.00 mol

Initial temperature = 390 K

Supplied power = -7.9 kW

Rate of heat = -1.1 kW

We need to calculate the time for compressor

Using formula of compression

\terxt{time for compression}=\text{time for half revolution}

\terxt{time for compression}=\dfrac{1}{2}\times T

\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}

Put the value into the formula

\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60

\terxt{time for compression}=0.2\ sec

We need to calculate the rate of internal energy

Using first law of thermodynamics

U=Q-W

\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}

Put the value into the formula

\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)

\dfrac{\Delta U}{\Delta t}=6.8\ kW

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}

\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}

Put the value into the formula

\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}

\Delta\theta=32.48^{\circ}

Hence, The temperature change per compression stroke is 32.48°.

6 0
3 years ago
A simple pendulum is made from a 0.54-m-long string and a small ball attached to its free end. The ball is pulled to one side th
Serga [27]

Answer:

0.37sec

Explanation:

Period of oscillation of a simple pendulum of length L is:

T = 2 π × √ (L /g)

L=length of string 0.54m

g=acceleration due to gravity

T-period

T = 2 x 3.14 x √[0.54/9.8]

T = 1.47sec

An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.

The ball will first have V(max) at T/4,

=>V(max) = 1.47/4 = 0.37 sec

3 0
3 years ago
Determine the magnitude and direction of the resultant force of the following free body diagram.
Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force (\vec F), in newtons, by vectorial sum:

\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j} (1)

\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}

\|\vec F\| \approx 599.923\,N

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)

Where \theta is the direction of the resultant force, in sexagesimal degrees.

\theta \approx 36.405^{\circ}

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

4 0
3 years ago
PLEASE Please HELP ME...
cupoosta [38]

Answer:

girl this easy ask yo teacher for help lol

7 0
3 years ago
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