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3 years ago

What can pressure be measured by?

Physics

1 answer:

SOVA2 [1]3 years ago

5 0

'Pressure' is (force) / (area).

The only choice with those units is #1 .

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Describe how the three methods of thermal energy transfer may take place within the iguana’s enclosure.

sasho [114]

Answer:

Heat can travel from one place to another in three ways: Conduction, Convection and Radiation. Both conduction and convection require matter to transfer heat. Conduction is the transfer of heat between substances that are in direct contact with each other. Thermal energy is transferred from hot places to cold places by convection. Radiation is a method of heat transfer that does not rely upon any contact between the heat source and the heated object as is the case with conduction and convection. Heat can be transmitted through empty space by thermal radiation often called infrared radiation.

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2 years ago

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B. 5 hours

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3 years ago

Franny drew a diagram to compare images produced by concave and convex lenses.

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3 years ago

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If measurements of gas are 75 L and 300 kIlopascals and then the gas is measured is second time and found to be 50 L, describe w

Ann [662]

Answer:

The pressure must have increased in the process

Explanation:

The State Equation for gasses reads: $P*V=n*R*T$

where P is the gas' pressure, V its volume, n the number of moles of gas, R the gas constant and T the temperature in degrees Kelvin.

If the temperature of the gas doesn't change in the described process, the right hand side of the equation stays the same. If that is the case, given that when the Volume of the gas diminishes from 75 liters to 50 liters, then the pressure must have increased to keep that product "P * V" constant:

$P_i*V_i=P_f*V_f\\75 *300=50*X\\X=\frac{75*300}{50} =450$

So the pressure must have gone up to 450 kilopascals.

3 0

3 years ago

Three resistors are wired in parallel with a battery. Two of the resistors have resistances of 38.7 Q/ and 89.5 Q. The current i

Lina20 [59]

Answer:

$214.9 \Omega$

Explanation:

The three resistors are connected in parallel: this means that the potential difference across each resistor is the same as the voltage of the battery. This can be calculated using the information about the $38.7 \Omega$ resistor: in fact, since we know its resistance and the current flowing through it (0.155 A), we can find the potential difference across this resistor, which is equal to the voltage of the battery:

$V=IR=(0.155 A)(38.7 \Omega)=6.0 V$

We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:

$R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega$

So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega$

4 0

3 years ago