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Anuta_ua [19.1K]
3 years ago
10

During a particular thunderstorm, the electric potential between a cloud and the ground is vcloud - vground = 1.5 x 108 v, with

the cloud being at the higher potential. what is the change in an electron's potential energy when the electron moves from the ground to the cloud?
Physics
1 answer:
lord [1]3 years ago
5 0
Voltage is potential energy per coulomb (J/C). So use the voltage and charge on an electron to get E=V•Q=1.5e8•1.602e-19=2.4e-11J
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nignag [31]

The position of the particle is given by:

x(t) = t³ - 12t² + 21t - 9

Differentiate x(t) with respect to t to find the velocity x'(t):

x'(t) = 3t² - 24t + 21

Differentiate x'(t) with respect to t to find the acceleration x''(t):

x''(t) = 6t - 24

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What is air masses? And 5 effect of it​
zhannawk [14.2K]

Answer:

an air mass is a volume of air defined by its temperature and water vapor content. Air masses cover many hundreds or thousands of miles, and adapt to the characteristics of the surface below them. They are classified according to latitude and their continental or maritime source regions. Colder air masses are termed polar or arctic, while warmer air masses are deemed tropical. Continental and superior air masses are dry while maritime and monsoon air masses are moist. Weather fronts separate air masses with different density (temperature and/or moisture) characteristics. Once an air mass moves away from its source region, underlying vegetation and water bodies can quickly modify its character.When winds move air masses, they carry their weather conditions (heat or cold, dry or moist) from the source region to a new region. When the air mass reaches a new region, it might clash with another air mass that has a different temperature and humidity. This can create a severe storm.

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5 0
3 years ago
A convex lens with a focal length of 10 cm is located 30 cm away from a
zloy xaker [14]

Answer: 15 cm

Explanation:

According to the Lens Equation we have the following:

\frac{1}{f}=\frac{1}{v}+\frac{1}{u} (1)

Where:

f=10 cm is the focal length

u=30 cm is the distance between the candle (the object) and the lens

v is the distance between the image and the lens

Isolating v:

v=\frac{uf}{u-f}} (2)

Solving:

v=\frac{(30 cm)(10 cm)}{30 cm-10 cm}} (3)

Finally:

v=15 cm This is where the image is located

8 0
2 years ago
A ball rolls 12m in 2.0s. What is the ball’s average velocity?
USPshnik [31]

Answer:

6 m/s

Explanation:

12m / 2s = 6 m/s

Hope that's the answer you seek.

5 0
3 years ago
8. A 40.0kg block of metal is suspended from a scale and is immersed in water. The dimensions of the block are 12.0cm x 10.0cm x
Alja [10]

Answer:

a.1017.9N

b.1029.7N

c.T=380.6N

d.11.8 N

Explanation:

(a) The absolute pressure at the level of the top of the block is

Po=atmospheric pressure

g=gravity

h1=height

rh0=density of water

P1=Po+rho*g*h1

1.013*10^5+1000(9.81)(0.05)

1.0179*10^5Pa

at the level of the bottom of the block we have

P2=Po+rho*g*h2

1.013*10^5+1000(9.81)(0.17)

1.0297* 10^5 Pa

the downward force exerted on the top by the water is

Ftop=P*A== × = 1.0179* 10^5* 0.100

= 1017.9 N

and the upward force the water exerts on the bottom of the block , which is the buoyant force

Fbot== × = 1.0297 10^5 * 0.100 m

= 1029.7 N

(b) The scale reading is the tension, T, in the cord supporting the block.

if the block is at equilibrium, then sum of vertical forces

EFy=T+Fbot-Ftop-mg=0

T=mg+Ftop-Fbot

T=40*9.81-(1029.7-1017.9)

T=380.6N

(c)  Archimedes principle state that, the buoyant force on the block equals the weight of the displaced water. Thus,

Buoyant force=rho *g*h

= = 1000* 0.100^2 * 0.120 m *9.80 m s=11.8 N

from the answer a Ftop-Fbot

1029.7-1017.9=11.8N, the same as the buoyant force

5 0
3 years ago
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