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katrin [286]
3 years ago
15

The lower atmosphere is mostly warmed by radiated heat from Earth's surface. However, water heats up and cools down more slowly

than land. Knowing this, which of the following statements is most likely true?
A. On a cloudy day, the air over a northern section of the ocean will be warmer than the air over a southern section of the ocean.

B. On a sunny day, the air over a lake will be cooler than the air over the bordering land.

C. On a cloudy day, the air over a southern section of the ocean will be warmer than the air over a northern section of the ocean.

D. On a sunny day, the air over a piece of land will be cooler than the air over a bordering lake.
Physics
2 answers:
mr_godi [17]3 years ago
7 0

Answer:

B

Explanation:

JulsSmile [24]3 years ago
6 0
The answer is B. On a sunny day, the air over a lake will be cooler than the air over the bordering land.
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If a proton were released from rest at the sphere's surface, what would be its speed far from the sphere?
balu736 [363]

Let the sphere is having charge Q and radius R

Now if the proton is released from rest

By energy conservation we can say

U = K

\frac{kQe}{R} = \frac{1}{2}mv^2

\frac{2kQe}{mR} = v^2

now take square root of both sides

v =\sqrt{\frac{2kQe}{mR}}

so the proton will move by above speed and

here Q = charge on the sphere

R = radius of sphere

k = 9 * 10^9


5 0
3 years ago
Read 2 more answers
8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
3 years ago
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Lorico [155]

Answer:2.6 h

Explanation:

Given

Total Trip distance=450 miles

Meeting starts after 10.8 hours

safe Fastest speed is  55 mi/h

so if he drives all the to the meeting with max speed then it takes =\frac{450}{55}=8.181 h

and total allowable time is 10.8

Therefore longest time he can spend over dinner is 10.8-8.181 \approx 2.6 hours

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