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garik1379 [7]
3 years ago
11

If you go to the beach on a hot summer day, the temperature of the sand is much higher than the temperature of the water. If we

assume the same amount of energy was supplied by the sun to both the sand and the water, does sand or water require more energy to raise its temperature?
Physics
2 answers:
guapka [62]3 years ago
7 0
Water has a higher specific heat so it requires more energy to heat than sand.
pentagon [3]3 years ago
7 0

Water requires more energy to raise its temperature than sand does.  In fact, of all the common substances that we see around us every day, water is one of the BEST at storing heat energy.

This is a big part of the reason why we use frozen water to cool our soda, instead of cold wood or cold steel balls.  

It's also a big part of the reason why we warm up the bed in the Winter with a hot water bag, instead of a bag of hot rocks or hot BBs.

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A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re
fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

PV=nRT

For a gas

P_{1}V_{1}=nRT_{1}

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

10\times V=nR\times286....(I)

When the temperature of the gas is increased

Then,

P_{2}V_{2}=\dfrac{n}{2}RT_{2}....(II)

Divided equation (I) by equation (II)

\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}

\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}

P_{2}=\dfrac{10\times368}{2\times286}

P_{2}= 6.433\ atm

P_{2}=6.4\ atm

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0
3 years ago
We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi
antoniya [11.8K]

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

Kc=510=1/2*Ic*\omega c^2

\omega c=\sqrt{510*2/Ic}

Where the inertia is given by:

Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2

Replacing this value:

\omega c=106.46rad/s

Speed of the block will therefore be:

V_b=\omega_c*R_c=106.46*0.15=15.969m/s

By conservation of energy:

Eo = Ef

Eo = 0

Ef = 510+1/2*m_b*V_b^2-m_b*g*h

So,

0 = 510+1/2*m_b*V_b^2-m_b*g*h

Solving for h we get:

h=16.67m

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3 years ago
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Answer: <u><em>A</em></u>

Explanation:

A pneumatic control system uses compressed air as a method of control for HVAC systems. ... Each senor responds to changes in temperature, humidity, and static pressure as examples, to provide feedback in a control loop to open or close the actuator to meet the control set point.

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