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horsena [70]
4 years ago
14

Water evaporates off lakes. Winds blow across the planet. Where does the energy come from for these and other weather processes?

Physics
2 answers:
Otrada [13]4 years ago
8 0

Answer:

B. Solar energy

Explanation:

The water cycle is driven primarily by the energy from the sun. This solar energy drives the cycle by evaporating water from the oceans, lakes, rivers, and even the soil. Other water moves from plants to the atmosphere through the process of transpiration.

lara [203]4 years ago
7 0

Answer: Solar Energy

Explanation:

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A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. How long does it take to come to rest? How many
Darya [45]

Answer:

t=45.7s

\alpha =116revolutions

Explanation:

Since we have given values of ω₀=32.o rad/s ,ω=0 and α=-0.700 rad/s² to find t we use below equation

w=w_{o}+at\\  0=(32.0rad/s)+(-0.700rad/s^{2} )t\\t=\frac{-32.0}{-0.700} \\t=45.7s

To find revolutions we use below equation

w^{2}=w_{o}^{2}+2a\alpha

Substitute the given values to find revolutions α

So

0=(32.0rad/s)^{2}+2(-0.700rad/s^{2} )\alpha  \\\alpha =\frac{(-32.0rad/s)^{2}}{2(-0.700rad/s^{2} )} \\\alpha =731rad

To convert rad to rev:

\alpha =(731rad)*(\frac{1rev}{2\pi rad} )\\\alpha =116revolutions

7 0
4 years ago
An object is 1.0 cm tall and its erect image is 5.0 cm tall. what is the exact magnification?
ikadub [295]
The exact magnification of the objects is calculated by dividing the cinema. We calculate it by diving the erect image size by the object size. From the given above, we find the exact magnification by dividing 5.0 cm by 1.0 cm. Thus, the answer would be 5. 
7 0
3 years ago
The tundra is located _______.
Leona [35]

Answer:

d: both north and south

Explanation:

8 0
3 years ago
Read 2 more answers
A tsunami covers a distance of 500km in 83.3 minutes. If the period of vibration of the molecules is 35.0 minutes, what is the w
marin [14]

Answer:

210079.798 m

Explanation:

pls give brainliest :)

8 0
3 years ago
A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui
GalinKa [24]

Answer:

n_l = 1.97

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

focal length of lens in liquid is

\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

                =\frac{n_{g} -n_{l}}{n_{l}}  [\frac{1}{(n_{g} - 1) f}

rearrange fron_l

n_l = \frac{n_g f_l}{f_l+f(n_g-1)}

n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}

n_l = 1.97

7 0
3 years ago
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