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3 years ago

The temperature of a sample of silver increased by 24.0 °C when 269 J of heat was applied. What is the mass of the sample?

Physics

1 answer:

statuscvo [17]3 years ago

3 0

Answer:

Mass of the silver will be equal to 46.70 gram

Explanation:

We have given heat required to raise the temperature of silver by 24°C is 269 J , so $\Delta T=24^{\circ}C$

Specific heat of silver = 0.240 J/gram°C

We have to find the mass of silver

We know that heat required is given by

$Q=mc\Delta T$ , here m is mass, c is specific heat of silver and $\Delta T$ is rise in temperature

So $269=m\times 0.240\times 24$

m = 46.70 gram

So mass of the silver will be equal to 46.70 gram

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An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10

STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

$P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}$

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi $(\frac{6894.76\ Pa}{1\ psi})$ = 379212 Pa

$\rho$ = density of water = 1000 kg/m³

Therefore,

$v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}$

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

$\frac{V}{t} = Av\\\\t =\frac{V}{Av}$

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = $\frac{\pi (0.015875\ m)^2}{4}$ = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = $[\frac{\pi (9.144\ m)^2}{4}][1.524\ m]$ = 100.1 m³

Therefore,

$t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\$

t = 18353.5 s = 305.9 min = 5.1 hours

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

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2 years ago

When the voltage across a steady resistance is doubled, the current?

natima [27]

I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.

Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know V and I use formula P = IV: P = IV = (100mA)(10V) = 1 W.

The next question is what I'm not sure about:

Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).

What I did initially was: P = IV = (100mA)(2V) = 2 W

But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."

So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.

P = IV = (200mA)(2V) = 4 W

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3 years ago

Which of the following is an example of chemical change?

Maru [420]

Answer:

Explanation:

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3 years ago

Read 2 more answers

PLZ HELP IM REALLY CONFUSED a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, the

erastovalidia [21]

Answer:

K.E = 100 J

Final P.E = 100 J

Explanation:

The kinetic energy of any object can be given by the following formula:

$K.E = (\frac{1}{2})mv^{2}$

where,

K.E = Kinetic Energy

m = mass of ball = 2 kg

v = speed of ball

Initially, v = 10 m/s. Therefore, the initial K.E is given as:

$K.E = (\frac{1}{2})(2\ kg)(10\ m/s)^{2}$

K.E = 100 J

Now, at the highest point the K.E of the ball becomes zero. because the ball stops for a moment at the highest point and its velocity becomes zero. So, from Law of Conservation of energy:

Initial K.E + Initial P.E = Final K.E + Final P.E

Initial P.E is also zero due to zero height initially.

K.E + 0 = 0 + Final P.E

Final P.E = 100 J

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3 years ago

The radius of Earth is . The average Earth-Sun distance is . How many Earths would fit between Earth and the Sun if they are sep

Verdich [7]

Answer:

The radius of the earth is 6,371 km.

The average Earth-Sun distance is 152.09 million km

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I didn't really understand the last part, but if you don't get a better answer please mark me as brainliest.

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2 years ago