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ANEK [815]
3 years ago
15

It has been a hot summer, so when you arrive at a lake, you decide to go for a swim even though it it nighttime. The water is co

ld. The next day, you go swimming again during the hottest part of the day, and even though the air is warmer the water is still almost as cold. Why?
(a) Water is fairly dense compared with many other liquids.
(b) Water remains in a liquid state for a wide range of temperatures.
(c) Water has a high bulk modulus.
(d) Water has a high specific heat.
Physics
1 answer:
nikitadnepr [17]3 years ago
8 0

Answer:

(d) Water has a high specific heat.

Explanation:

At night, when the temperature of earth goes down due to loss of heat , the temperature of water is lost slowly and temperature of land is lost fast because of high specific heat of water . Water loses as well as gains temperature comparatively slowly due to its high specific heat .

During daytime when earth gains heat , the temperature of land rises more rapidly than water so water appears cool even during daytime when land becomes  hotter . It is also due to high heat holding capacity of water or due to high specific heat of water .

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It would exert the same back right?
3 0
3 years ago
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

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So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
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Here when car in front of us applied brakes then it is slowing down due to frictional force on it

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now it is given that

\mu = 0.868

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so here we have

a = -0.868 * 9.81

a = -8.52 m/s^2

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4 0
3 years ago
Why won’t anyone help me please anybody help me I really need help .
irga5000 [103]

Answer:

1➡️ this is the method of decomposition

2➡️ H2 and O2

3➡️ b

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8 0
2 years ago
The Coefficient of kinetic friction between the tires of your car and the roadway is \"μ\". (a) If your initial speed is \"v\" a
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We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation           -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
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3 0
3 years ago
Read 2 more answers
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