<span>They would feel that the water is cold.
</span> The atmosphere is heated both by the Sun and by the Earth's surface. Water radiates heat differently than land, so the air temperature over the ocean is usually different than the air temperature over land. <span>
The difference in air temperature over land compared to over water causes convection currents in the atmosphere. How would a person at the beach experience these convection currents?
</span>They would feel that the water is cold.
NOT:
They would feel the heat of the Sun.
They would feel that the sand is hot.
<span>They would feel wind as the air moves.</span>
Answer:
7.16 m /s
Explanation:
The depth of the small pipe attached with the side wall of tank from the surface of water
h =( 3.1 - .48 )m
= 2.62 m
velocity of flow of water= √ 2 g h
= √ 2 x 9.8 x 2.62
= 7.16 m /s
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Answer:
the speed of the block when it reaches point B is 14 m/s
Explanation:
Given that:
mass of the block slides = 1.5 - kg
height = 10 m
Force constant = 200 N/m
distance of rough surface patch = 20 m
coefficient of kinetic friction = 0.15
In order to determine the speed of the block when it reaches point B.
We consider the equation for the energy conservation in the system which can be represented by:
v = 14 m/s
Thus; the speed of the block when it reaches point B is 14 m/s
Answer:
<em>The difference in pressure between the external air pressure, and the internal air pressure of the middle ear.</em>
Explanation:
First of all, we should note that pressure decreases with height and increases with depth. The air within the middle ear (between the ear drum and the Eustachian tube) adjusts itself to respond to the atmospheric pressure, or when we yawn. At a high altitude like on the hill, the air pressure in the middle ear, is fairly low (this is to balance the low air pressure at this height). While riding down the hill quickly, there is little time for the air pressure in the ear to readjust itself to the increasing external air pressure, causing the external air to push into the ear drum. Along the way, the air within the middle ear is adjusted by the opening of the Eustachian tube, allowing more air into the space in the middle ear to balance the external air pressure. This readjustment causes the ear to pop.
Answer:
A. Final pressure P2
P2/P1 = (T2/T1)^n/n-1
P1 = 4bar
T1 = 438K
T2 = 300K
Polytropic index, n, = 1.3
P2 = 4 (300/438)^1.3/1.3-1
P2 = 4 (300/438)^4.333
P2 = 4 * 0.19400
P2 = 0.776bar.
B. The work done is;
W = mR/ n-1 (T1 -T2)
Where, R = 0.1889kJ/kg.K, m = 1
W = 1 * 0.1889/ 1.3-1 * (438-300)
W = 86.89kJ/kg.
C. The heat transfer, Q
Q = W + ΔU
Q = W + mCv(T2-T1), where Cv of nitrogen is 0.743kj/kgk
Q = 86.89 + 1 * 0.743 (300-438)
Q = 86.89 + (-102.534)
Q = -15.644kJ/K
Q = 15.64kJ/K
