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fredd [130]
2 years ago
14

A balloon was filled to a volume of 2.50 l when the temperature was 30.0∘c. What would the volume become if the temperature drop

ped to 11.0∘c.
Physics
1 answer:
Scilla [17]2 years ago
4 0

Answer:

2.34 L

Explanation:

Assuming the pressure inside the balloon remains constant, then we can use Charle's law, which states that for a gas kept at constant pressure, the ratio between the volume of the gas and its temperature remainst constant:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where in this problem we have:

V_1 = 2.50 L is the initial volume

V_2 is the final volume

T_1 = 30.0^{\circ}C+273 = 303 K is the initial temperature

T_2 = 11.0^{\circ}C+273 = 284 K is the final temperature

Substituting into the equation and solving for V2, we find the final volume:

V_2 = \frac{V_1 T_2}{T_1}=\frac{(2.50 L)(284 K)}{303 K}=2.34 L

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-- From the point of view of an observer on the sun, or on any other planet,
the moon's orbit is an ellipse with the sun at one focus, and perturbed by being
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2 years ago
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A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the vertical velocity of t
zhuklara [117]

Answer:

<em>You would use the kinematic formula:</em>

    \Delta y=V_{0y}\times t-g\times t^2/2

Explanation:

The upwards vertical motion is ruled by the equation:

        y=y_0+V_{0y}\times t-g\times t^2/2

Where:

       y \text{ is the position at the time }t:y=0.1m

       y_0\text{ is the initial position: }y_0=0

       t=2s

       g\text{ is the gravitational acceleration: }\approx 9.8m/s^2

       V_{0y}\text{ is the initial vertical velocity}

Naming Δy = y - y₀, the equation becomes:

      \Delta y=V_{0y}\times t-g\times t^2/2

Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.

7 0
1 year ago
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A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

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Kindly check the diagram in the attached image below.

5 0
1 year ago
A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C
Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 \frac{J}{kg k}

The heat of vaporization is =  2256 \frac{KJ}{kg}

Mass = 0.030 kg

Energy needed to heat water to boiling point =  m c ( T_{2} - T_{1} )

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy (E_{1}) = 10.08 KJ

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Thus the total energy needed E =  E_{1} + E_{2}

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

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Explanation:

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\omega=2\sqrt{2\pi \alpha}

Hence, this is the required solution.

6 0
2 years ago
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