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2 years ago

a hot piece of copper is placed in an insulated cup. what is the final temperature of the water and copper?

Physics

1 answer:

slavikrds [6]2 years ago

8 0

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

22°C

Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

Mass of copper (Mc) = 0.5 Kg

Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

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Any object that just fits into your visual field has this angular size. Do all objects that you see this this angular size have

Nataly_w [17]

Answer:

No the angular and linear size will not be same for different objects bearing the same angular size and is dependent on the point of observation. The greater the distance between the observer and the object, the greater the difference between the angular and linear size.

Explanation:

The angular size is the ratio of two lengths namely

$Angular \, Size=\frac{Linear Size}{Distance from the object}$

Here if the distance from the object is unity, the angular and linear size will be similar. However if the distance of observation is very large, the angular size for large bodies with large linear size will be very small.

An example in this regard is Sun.

The linear size of Sun, (the diameter) is 1.3927 million km. Which is very large. However as it is very far from earth, 147.44 million km, the angular size is very small.

it is given as

$Angular \, Size=\frac{1.39}{147.44}=0.009 rad$

Now the same angular size can be of a tennis ball having a diameter of 10 cm , placed at around 10.6 m away.

$Angular \, Size=\frac{0.1}{10.6}=0.009 rad$

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3 years ago

An emu moving with constant acceleration covers the distance between two points that are 92 m

nasty-shy [4]

Answer:

a) $V_{o}=14.30 m/s$

b) $a=-0.046 m/s^{2}$

Explanation:

The complete question is written below:

An emu moving with constant acceleration covers the distance between two points that are 92 m apart in 6.5s. Its speed as it passes the second point is 14 m/s. What are (a) its speed at the first point and (b) its acceleration?

Since we are talking about constant acceleration, we can use the following equations:

$d=x-x_{o}=(\frac{V_{o}-V}{2})t$ (1)

$V=V_{o}+at$ (2)

Where:

$d=92 m$ is the distance between the two points

$V_{o}$ is the velocity of the emu at the first point

$V=14 m/s$ is the velocity of the emu at the second point

$t=6.5 s$ is the time it takes to the emu to cover the distance $d$

$a$ is the emu's constant acceleration

Knowing this, let's begin with the answers:

<h2>a) Speed at the first point</h2>

In this situation wi will use equation (1):

$d=(\frac{V_{o}-V}{2})t$ (1)

Finding $V_{o}$ :

$V_{o}=\frac{2d}{t}-V$ (3)

$V_{o}=\frac{2(92 m)}{6.5 s}-14 m/s$ (4)

$V_{o}=14.30 m/s$ (5)

<h2>b) Emu's acceleration</h2>

Now we will substitute (5) in equation (2):

$14 m/s=14.30 m/s+a(6.5 s)$ (6)

Finding $a$ :

$a=-0.046 m/s^{2}$ (7) This means the emu is decreasing its speed at a constant rate.

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3 years ago

Dump Tower is 96 stories tall. A small, 1.2-kg object is dropped over the side of the roof of the tower and accelerates toward t

laiz [17]

Answer:

6.0 s

98 m/s

Explanation:

The radius of the planet is much bigger than the height of the tower, so we will assume the acceleration is constant. Neglect air resistance.

Acceleration due to gravity on this planet is:

a = GM / r²

a = (6.67×10⁻¹¹ m³/kg/s²) (2.7 × 1.48×10²³ kg) / (1.7 × 750,000 m)²

a = 16.4 m/s²

The height of the tower is:

Δy = 96 × 3.05 m

Δy = 293 m

Given v₀ = 0 m/s, find t and v.

Δy = v₀ t + ½ at²

(293 m) = (0 m/s) t + ½ (16.4 m/s²) t²

t = 6.0 s

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (16.4 m/s²) (293 m)

v = 98 m/s

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3 years ago

What happens when a electron and a proton get close together?

Vladimir79 [104]

Answer:

it leads to attraction, only because the total energy of the electron is negative. Therefore the electron moves closer to the proton rather than farther from it an attractive force.... Electrons are negatively charged and protons are positively charged.

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3 years ago

What causes atoms to bond together<br> ?!?

lisov135 [29]

Answer:

electrostatic attraction

Explanation:

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3 years ago

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