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3 years ago

1. How much heat must be absorbed by 375 grams of water to raise its

Physics

1 answer:

antiseptic1488 [7]3 years ago

5 0

Answer:

39225J

Explanation:

Given parameters:

Mass of water = 375grams of water

Change in temperature = 25°C

Specific heat capacity of water = 4.184J/g°C

Unknown:

Amount of heat absorbed = ?

Solution:

To solve this problem, we use the expression below:

H = m c Ф

H is the heat absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

Insert the parameters and solve;

H = 375 x 4.184 x (25) = 39225J

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4 years ago

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The drawing shows three different resistors in two different circuits. the battery has a voltage of v = 23 v, and the resistors

vesna_86 [32]

Missing picture in the text. I've found it here:
https://www.flickr.com/photos/[email protected]/6791933847/in/photostream

Solution:

Circuit 1)
The three resistors are in series. So the equivalent resistance of the circuit is
$R_{eq}=R_1+R_2+R_3 =50.0\Omega+25.0 \Omega+10.0 \Omega=85.0 \Omega$
So the current flowing through each resistor is the same, and it is given by Ohm's law:
$I= \frac{V}{R_{eq}}= \frac{23 V}{85.0 \Omega}=0.27 A$

And the voltage drop across each resistor is given by Ohm's law as well:
$V_1 = I R_1 = (0.27 A)(50.0 \Omega)=13.5 V$
$V_2 = I R_2 = (0.27 A)(25.0 \Omega)=6.8 V$
$V_3 = I R_3 = (0.27 A)(10.0 \Omega)=2.7 V$

Circuit 2)
The three resistors are in parallel, so their equivalent resistance is given by
$\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{50\Omega}+ \frac{1}{25 \Omega}+ \frac{1}{10 \Omega}= \frac{7}{50 \Omega}$
$R_{eq}= \frac{50}{7} \Omega=7.1 \Omega$

The voltage drop across each resistor is the same, and it is equal to the voltage of the battery:
$V_1 = V_2 = V_3 = 23 V$

while the current across each resistor is given by Ohm's law:
$I_1 = \frac{V}{R_1}= \frac{23 V}{50 \Omega}=0.46 A$
$I_2 = \frac{V}{R_2}= \frac{23 V}{25.0 \Omega}=0.92 A$
$I_3 = \frac{V}{R_3}= \frac{23 V}{10 \Omega}=2.3 A$

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3 years ago

A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.

dedylja [7]

Answer:

F = 768 N

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, $\dfrac{dm}{dt}=240\ kg/min = 4\ kg/s$

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

$F=\dfrac{dp}{dt}$

$F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$

Since, the speed is constant,

$F=m\dfrac{dv}{dt}$

$F=v\dfrac{dm}{dt}$

$F=3.2\times 240$

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

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4 years ago

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4 years ago

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