calculate the resistance of 50m length of wire of cross sectional area 0.01 square mm and of resistivity 5*10 to the power minus

1. Which of the following is not an acceptable Physical Education activity?

Dmitrij [34]

The answer is A

Explanation:
Vacuuming doesn’t involve a lot of physical movements.

3 0

3 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

If we assume that, in the African savanna, the predator population remains constant, what other, nonmobile, factor would influen

musickatia [10]

On usa test prep the answer is c

7 0

3 years ago

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Explore the role of friction when creating safety ramps and escalators. "Why is it important for people and object to move slowl

faust18 [17]

The role of friction is of great importance when creating safety ramps and escalators because with the help of friction things move.

<h3>Why is it important to move objects slowly on ramps and escalator?</h3>

It is important to move objects slowly on ramps and escalator because the ramps and escalator moves object in the opposite direction of gravity. If we did not move objects slowly, then the objects or a person get damaged.

So we can conclude that the role of friction is of great importance when creating safety ramps and escalators because with the help of friction things move.

Learn more about friction here: brainly.com/question/24338873#SPJ1

3 0

2 years ago

A boy standing throws a penny horizontally at 7.25 m/s out of the window of his apparent buliding. If the window is 10.0 m above

Ksivusya [100]

Answer:

Explanation:

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + $\frac{1}{2}at^2$ and plugging in from the y-dimension info:

$-10.0=0t+\frac{1}{2}(-9.8)t^2$ which simplifies to

$-10.0=-4.9t^2$ so

$t=\sqrt{\frac{-10.0}{-4.9} }$ which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in the x-stuff:

Δx = $7.25(1.4)+\frac{1}{2}(0)(1.4)^2$ Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:

Δx = 7.25(1.4) + 0 so

Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).

6 0

3 years ago

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