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vesna_86 [32]
3 years ago
10

A balloon is filled at sea level on a spring day (temperature 20°C) to a volume of 0.500 m3 and then brought at op of Mount Ever

est the same day (temperature -20°C). How does its volume change? What is the percent change?
Physics
1 answer:
blagie [28]3 years ago
8 0

Answer:

Volume increases

Explanation:

The balloon when filled at sea level being comparatively close to the center of the earth will have higher pressure due to the influence of gravity and when this balloon is taken to the top of the mountain being away from the center of earth, it will experience a lesser pressure due to low gravity where the amount of force exerted by the air on the object is lesser as compared to to that at the sea level.

Therefore, there will be an increase in volume of the balloon as there is expansion of air on the inside of the balloon as a result of low pressure.

There is not  much effect of temperature at both the sea level and the mountain top as the temperature does not impart any energy to the air molecules so as to decrease the volume.

Therefore,there is an increase in the volume of the balloon at the top of the mountain.

Percentage change in volume is given by:

% change = \farc{V - 0.5}{0.5}times 100

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The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

Where,

d_g = Depth of glass

n_w = Refraction index of water

n_g = Refraction index of glass

n_{air} = Refraction index of air

d_w = Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})

d'w = 4.041cm

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

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The electric current running through the wire coil in an electric motor exerts force directly onto
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C) A powerful magnet
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A 0.20 kg mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured
xxMikexx [17]

Answer:

b. 0.20 m/s.

Explanation:

Given;

initial mass, m = 0.2 kg

maximum speed,  v = 0.3 m/s

The total energy of the spring at the given maximum speed is calculated as;

K.E = ¹/₂mv²

K.E = 0.5 x 0.2 x 0.3²

K.E = 0.009 J

If the mass is changed to 0.4 kg

¹/₂mv² = K.E

mv² = 2K.E

v = \sqrt{\frac{2K.E}{m} } \\\\v =  \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s

Therefore, the maximum speed is 0.20 m/s

4 0
3 years ago
the electrical resistance of copper wire varies directly with its length and inversely with the square of the diameter of the wi
leonid [27]

Answer:24.47

Explanation:

Given

L_1=30 m

d_1=3 mm

R_1=25 \Omega

L_2=40 m

d_2=3.5 mm

we know Resistance R=\frac{\rho L}{A}

Where R=resistance

\rho =resistivity

L=Length

A=area of cross-section

A=\frac{\pi d^2}{4}

Thus R\propto \frac{L}{d^2}

therefore

R_1\propto \frac{L_1}{d_1^2}--------1

R_2\propto \frac{L_2}{d_2^2}--------2

divide 1 and 2 we get

\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{d_2^2}{d_1^2}

\frac{R_1}{R_2}=\frac{30}{40}\times \frac{3.5^2}{3^2}

R_2=25\times 0.734\times 1.33

R_2=24.47 \Omega

8 0
3 years ago
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