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3 years ago

Warm air is more dense than cool air. Warm air is less dense than cool air Warm air and cool air have the same density

Physics

1 answer:

Wewaii [24]3 years ago

7 0

Answer:

You are exactly right. The molecules in hot air are moving faster than the molecules in cold air. Because of this, the molecules in hot air tend to be further apart on average, giving hot air a lower density. That means, for the same volume of air, hot air has fewer molecules and so it weighs less

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Scenario A: 120 J of work is done in 6 s. Scenario B: 160 J of work is done in 8 s. Scenario C: 200 J of work is done in 10 s. W

hodyreva [135]

Using the equation P = W/t to solve your problem .

Thus the answer is all of them use the same amount of power. 20 J.

8 0

3 years ago

Read 2 more answers

A plastic rod has been bent into a circle of radius R = 8.00 cm. It has a charge Q1 = +2.70 pC uniformly distributed along one-q

vovikov84 [41]

Answer:

Part a)

$V = -1.52 V$

Part b)

$V = -1.16 V$

Explanation:

Part a)

Electric potential is a scalar quantity

so here we can say that total potential due to a ring on its center is given as

$V = \frac{kQ}{R}$

here we know that

$Q = Q_1 - 6Q_1$

$Q = - 5 Q_1$

$Q_1 = 2.70 pC$

now we have

$V = \frac{(9\times 10^9)(-5\times 2.70 \times 10^{-12})}{0.08}$

$V = -1.52 V$

Part b)

Potential on the axis of the ring is given as

$V = \frac{kQ}{\sqrt{r^2 + R^2}}$

$V = \frac{(9\times 10^9)(-5\times 2.70\times 10^{-12})}{\sqrt{0.08^2 + 0.0671^2}}$

$V = -1.16 V$

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3 years ago

A basketball player standing under the hoop launches the ball straight up with an initial velocity of v₀ = 3.25 m/s from 2.5 m a

matrenka [14]

Answer: 0.53m

Explanation:

According to the equation of motion v²= v₀²+2as

Since the body is launched upward, the final velocity at the maximum height will be "zero" since the body will momentarily be at rest at the maximum height i.e v = 0

Initial velocity given (v₀) = 3.25 m/s

The body is also under the influence of gravity but the acceleration due to gravity will be negative being an upward force (a = -g) and the distance (s) will serve as our maximum height (h)

The equation of motion will.now become

V = v₀² -2gh

Where v = 0 v₀ = 3.25m/s g = 10m/s h = ?

0 = 3.25² - 2(10)h

0 = 10.56 - 20h

-10.56 = -20h

h = 10.56/20

h = 0.53m

Therefore, the maximum height, h (in meters), above the launch point that the basketball will achieve is 0.53m

6 0

4 years ago

Pulleys and Belts. In a pulley system, pulley A is moving at 1500 rpm and has a diameter of 15 in. Three pulleys, B, C, and D, a

NeTakaya

For this question I would use the similarities between pulley A, to each pulley of the question.

<span>1. A speed of 1750 rpm is required when the drive belt is connected to pulley B. What is the diameter of pulley B?</span>

15inches / 1500rpm = (x<span>) inches / 1750rpm</span>

(x) = 15 inches * 1750 rpm / 1500rpm

(x) = 17.5 inches

2.A speed of 2000 rpm is required when the drive belt is connected to pulley C. What is the diameter of pulley C?

15inches / 1500rpm = (x) inches / 2000rpm

(x) = 15 inches * 2000 rpm / 1500rpm

(x) = 20 inches

3. A speed of 3250 rpm is required when the drive belt is connected to?

15inches / 1500rpm = (x) inches / 3250rpm

(x) = 15 inches * 3250 rpm / 1500rpm

(x) = 32.5 inches

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4 years ago

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Masses M1 and M2 are initially separated by a distance Ra. Mass M2 is now moved further away from mass M1 such that their final

Komok [63]

Answer:

A. Wab > 0.

Explanation:

Work done will be equal to change in gravitational energy in the whole process .

When Masses M₁ and M₂ are initially separated by a distance Ra

their gravitational energy

= - G M₁ M₂ / Ra

When mass M₂ is now moved further away from mass M₁ such that their final separation is Rb

their gravitational energy

= - G M₁ M₂ / Rb

Change in gravitational energy

= - G M₁ M₂ / Rb - ( - G M₁ M₂ / Ra )

= G M₁ M₂ / Ra - - G M₁ M₂ / Rb

= G M₁ M₂ ( 1 / Ra - Rb )

As Ra < Rb

1/Ra > 1 / Rb

So change in gravitational energy is positive

Wab > 0.

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3 years ago