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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 speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

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Final speed of brick as it hits the ground; v =  \ ?

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

v^2 = u^2 + 2gh

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

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2 years ago
Which seismic waves are generally the last to arrive after an earthquake?
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A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate
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Answer:

N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons

Explanation:

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Q_{total}/2=(3.8*10^{-6}C+1.9C)/2=0.9500019C\\

The plate and the rod must have Q_{total}/2\\. So the charge transferred from the plate to the rod is:

Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C\\

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N_{electrons}=Q_{transfered}/q_{electron}=-0.9499981C/(-1.6*10^{-19}C)=5.94*10^{18}electrons

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In your own words, explain what conservation of energy means. Also, give an example of the conservation of energy using somethin
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A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
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Answer:

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Explanation:

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x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

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