A copper block is removed from a 310C oven and dropped into 1.10kg of water at 23.0C. The water quickly reaches 33.0C and then remains at that temperature.What is the mass of the copper block? The specific heats of copper and water are 385 J/(kg?K) and 4190 J/(kg?K) respectively. Express your answer to three significant figures and include the appropriate units.

Answers

Answer 1
Answer:

Answer:

0.432kg or 432g

Explanation:

Copper block is removed from 310C oven.

Mass of water = 1.10kg

Initial temperature of water (θ1) = 23.0C

Final temperature of water (θ2 ) = 33.0C

Specific heat of copper= 385

Specific heat of water = 4190

Let M (copper) be the mass of copper

The amount of heat lost will be absorbed by water

Q (net) = Q(water) + Q(copper) = 0 …………………(1)

Q = MCθ

For water,

Q(water) = M(water) * C(water) * (θ2-θ1)...........(2)

= 1.1 * 4190 * (33.0 -23.0)

= 1.1 *4190 * 10

= 46090J

For copper,

Q(copper) = M(copper) * C(copper) * (33.0 - 310)...........(3)

from equation 1,

Q(copper) = -Q(water)

M(copper) * C(copper) * (33.0 - 310) = - 46090

M(copper) * 385 * (-277) = -46090

M(copper) * 106645 = -46090

M(copper) = -46090 / -106645

M(copper) = 0.432kg

M(copper) = 432g

The mass of copper =432g

Answer 2
Answer:

Final answer:

The mass of the copper block is 3.69kg.

Explanation:

To find the mass of the copper block, we can use the equation q1 = q2. The heat gained by the water is given by q1 = mcΔT, where m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature. The heat lost by the copper block is given by q2 = mcΔT, where m is the mass of the copper block, c is the specific heat of copper, and ΔT is the change in temperature. Since the water reaches a final temperature of 33.0°C, the change in temperature is ΔT = 33.0°C - 23.0°C = 10.0°C.

Using the equation q1 = q2, we have mcΔT = mcΔT. Rearranging the equation to solve for the mass of the copper block, we get m = (m1c1ΔT2) / (c2ΔT1), where m1 is the mass of the water, c1 is the specific heat of water, ΔT2 is the change in temperature of the copper block, c2 is the specific heat of copper, and ΔT1 is the change in temperature of the water. Substituting the given values, we have m = (1.10kg * 4190 J/(kg?K) * 10.0°C) / (385 J/(kg?K) * 10.0°C), which simplifies to m = 3.69kg.

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A 1.0 kg ball is thrown into the air with an initial velocity of 30 m/s. What is the kinetic energy of the ball?

Answers

Answer:

If this helps, to work out the kinetic energy the formula is K (stands for kinetic energy) = 1/2 (half of) mass and velocity. So the answer is 450 joules.

Explanation:

A fastball pitcher can throw a baseball at a speed of 40 m/s (90 mi/h). (a) Assuming the pitcher can release the ball 16.7 m from home plate so the ball is moving horizontally, how long does it take the ball to reach home plate? (b) How far does the ball drop between the pitcher’s hand and home plate?

Answers

Answer:

a) 0.4175 seconds

b) 0.854 m

Explanation:

t = Time taken

u = Initial velocity = 40 m/s in x direction

v = Final velocity

s = Displacement = 16.7 m

a = Acceleration due to gravity = 9.81 m/s²

a) Time = Distance / Speed

\text{Time}=(16.7)/(40)=0.4175\ s

Time taken by the ball to reach home plate is 0.4175 seconds

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The ball drops by 0.854 m between the pitcher’s hand and home plate.

Which of the following does not represent a mechanical wave?sound
waves in the ocean
waves in a wheat field
light

Answers

Answer:

The other person who answered is incorrect- the real answer is light.

Explanation:

  A mechanical wave is a wave that requires a medium. A medium is any substance that has a defined set of physical properties and can transmit energy. A medium can be a solid, liquid, or gas. Some examples of a medium include an iron bar, the ground, water, lava, air, and helium in a balloon.

  The answer is not sound because sound travels through waves through the air. If you were to put a source of sound in a container that contains no air, you would not be able to hear the sound through the container. Since sound has a wave and a medium, it is a mechanical wave.

  The answer is not waves in the ocean because, obviously, this is a wave. Its medium is the water. Since it has a wave and a medium, it is a mechanical wave.

  The answer is not waves in a wheat field because, again, this is a wave. The medium is the air. Since it has a wave and a medium, it is a mechanical wave.

  The answer is light. Light travels through electromagnetic waves. However, it does not require a medium. Light can travel not only through air and solid materials, but also through the vacuum of space. Since light does not require a medium, it is not a mechanical wave.

I hope this helps !! :)

ocean waves because its not artificial

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c) tropic of Capricorn
d) tropic of Cancer
e) South pole

Answers

The equator does.  There, the noontime sun is never more than 23.5 degrees
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An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.5 x 10⁻⁷ C/m², and the plates are separated by a distance of 1.7 x 10⁻² m. How fast is the electron moving just before it reaches the positive plate?

Answers

The speed of the electron before reaching the positive plate is 1.30 * 10^(7)\ m / s

Explanation:

As per Gauss law of electro statistics, the electric field generated by a capacitor is directly proportional to the surface charge density of the plate and inversely proportional to the dielectric constant. In simple words, the electric field is proportional to the surface charge density.  So,  

    \text {Electric field}=(\sigma)/(\varepsilon_(0))

And then from the second law of motion, F=m * acceleration

So acceleration exerted by the electrons will be directly proportional to the force exerted on them and inversely proportional to the mass of the electron.

        Acceleration =(F)/(m)

Since force is also calculated as product of charge with electric field in electrostatic force,

       \text {Acceleration}=(q E)/(m)=(q \sigma)/(m \varepsilon_(0))

So, the charge of electronq=1.6 * 10^(-19)\ \mathrm{C}, \sigma=\text { Charge per unit area }=2.5 * 10^(-7)\ \mathrm{C} / \mathrm{m}^(2)

m is the mass of electron which is equal to 9.11 * 10^(-31)\ \mathrm{kg}

\varepsilon_(0)=8.85 * 10^(-12)\ \mathrm{Nm}^(2) \mathrm{C}^(-2)

Then,

    \text { Acceleration }=(1.6 * 2.5 * 10^(-19) * 10^(-7))/(9.11 * 8.85 * 10^(-31) * 10^(-12))=(4 * 10^(-19-7))/(80.62 * 10^(-31-12))

   \text { Acceleration }=0.0496 * 10^(-19-7+31+12)=0.0496 * 10^(17)\ \mathrm{m} / \mathrm{s}^(2)

So the acceleration of the electron in the capacitor will be 4.96 * 10^(15) m / s^(2)

Then, the velocity can be observed from the third equation of motion.

    v^(2)=u^(2)+2 a s

As u = 0 and s is the distance of separation between two plates.

   \begin{array}{c}v^(2)=0+\left(2 * 4.96 * 10^(15) * 1.7 * 10^(-2)\right) \nv^(2)=16.864 * 10^(15-2)=16.864 * 10^(13)=1.684 * 10^(14)\end{array}

Thus, v=\sqrt{\left(1.68 * 10^(14)\right)}=1.30 * 10^(7)\ m/s

So, the speed of the electron before reaching the positive plate is 1.30 * 10^(7) \mathrm{m} / \mathrm{s}.

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Answers

A. Mesosphere; exosphere
your answer to the question is a.