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3 years ago

Where is the switch located on this diagram?

Physics

1 answer:

Fed [463]3 years ago

6 0

For this case, the switch is located at point B of the diagram.
Remember that point D is the universal symbol for resistance.
In A what you have is a source of power and in C what you have is a cable.
Therefore, the answer for this case is B.

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A sample of gas in a syringe releases heat to its surroundings while the

blsea [12.9K]

Answer: b

Explanation:

When heat is released by the system i.e. system loses heat. So, we take it as negative -Q

When the work is done on the system then it is considered as negative work on the system i.e. -W

In this case, the plunger is pulled out, and work is done on the system. So, we take work as negative work -W

Correct option is b

3 0

3 years ago

Read 2 more answers

A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following fun

snow_lady [41]

The correct answer would be 1.375 < t < 3 i hope this helps anyone

8 0

3 years ago

Read 2 more answers

A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

7 0

3 years ago

How do its eight arms help an octopus obtain food

Andrews [41]

The more arms it has the less of a chance the prey has to swim away.

5 0

3 years ago

When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl

zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L = ρ*I / S

then

(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

S₁ = 0.25*π*D² and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒ I₂ = 4*ρ₁*I₁ / ρ₂

⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒ I₂ = 21.13 mA

5 0

3 years ago