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IRINA_888 [86]
3 years ago
5

State ohm’s law Q.34.7

Physics
1 answer:
defon3 years ago
4 0

Ohm's law states that: V=RI

Explanation:

Ohm's law states that in a conductor, the potential difference across the conductor is directly proportional to the current flowing through it. Mathematically,

V \propto I

where

V is the potential difference

I is the current

The constant of proportionality is called resistance (R), and it gives a measure of "how much the conductor opposes" to the flow of current. Therefore Ohm's law can be rewritten as

V=RI

where R is the resistance. By rewriting the equation as

I=\frac{V}{R}

we see that the larger the resistance, the lower the current in the conductor.

Learn more about Ohm's law:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

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A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?
allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side.  See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

F_g = m \cdot g

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3

The density of water is 997 kg/m^3, so the mass of the barrel is:

m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg

and so the weight is

F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N

and so the pressure is

P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.  

7 0
3 years ago
Now find the electromotive force E2(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1. Remembe
lions [1.4K]

Answer:

E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)

Explanation:

Consider two solenoids that are wound on a common cylinder as shown in fig. 1. Let the cylinder have radius 'ρ' and length 'L' .

No. of turns of solenoid 1 = n₁

No. of turns of solenoid 1 = n₂

Assume that length of  solenoid is much longer than its radius, so that its field can be determined from Ampère's law throughout its entire length:

\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= \mu_{o}I

We will consider the field that arises from solenoid 1, having n₁  turns per unit length. The magnetic field due to solenoid 1 passes through solenoid 2, which has n₂ turns per unit length.

Any change in magnetic flux from the field generated by solenoid 1 induces an EMF in solenoid 2 through Faraday's law of induction:

\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= -\frac{d}{dt} \phi _{M}(t)

Consider B₁(t) magnetic feild generated in solenoid 1 due to current I₁(t)

Using:

                                  B_{1}(t) =\mu _{o} nI(t)\\ --- (2)

                           

Flux generated due to magnetic field B₁

                      \phi _{1}(t) = \oint \overrightarrow {B_{1}}.dA\\ ---(3)

area of solenoid = A = \pi \rho^{2}

substituting (2) in (3)

                       \phi _{1}(t) = \mu_{o} \pi \rho^{2} n_{1}I_{1}(t) ----(4)

We have to find electromotive force E₂(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1, i.e.

                       E_{2} (t) = -n_{2}L\frac{d \phi_{1}}{dt} ---- (5)

substituting (4) in (5)

E_{2} (t) = -n_{2}L\frac{d }{dt}(\mu_o} \pi \rho^{2} n_{1}I_{1}(t))\\E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)

5 0
3 years ago
Can someone list 6 advantages of renewable energy.
igor_vitrenko [27]

Answer:

Solar energy,  Wind energy,  Hydro energy, Tidal energy, Geothermal energy,  Biomass Energy.

Explanation:

I hope that it helps you...

4 0
3 years ago
9. You drive a car from Milwaukee to Chicago, which is a distance of 150km and it takes you 95
Katena32 [7]

For this case we have to, by definition:

v = \frac {x} {t}

Where:

v: It's the velocity

x: It is the distance traveled

t: It is the time spent

1 hour equals 60 minutes and 1 minute equals 60 seconds.

On the other hand:

1 kilometer equals 1000 meters

So:\frac {150} {95} \frac{km} {min} * \frac {1} {60} \frac {min} {sec} * \frac {1000} {1} \frac {m} {km} = \frac {150 * 1000} {95 * 60} \frac {m} {sec} = \frac {15000} {5700} \frac {m} {sec} = 2.63 \frac {m} {sec}

On the other hand:\frac {150} {95} \frac{km} {min} * \frac {60} {1} \frac {min} {hr} = \frac {150 * 60} {95} \frac {km} {h} = 94.74 \frac {km} {h}

ANswer:

2.63 \frac {m} {sec}\\94.74 \frac {km} {h}

8 0
3 years ago
Velocity ratio of a machine is 4 what does it mean​
Jlenok [28]

Answer:

It means that the velocity of the machine is 4

3 0
3 years ago
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