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Sophie [7]
3 years ago
10

A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(

kg⋅K). The copper and water are initially at different temperatures. After a sufficiently long time, the copper and water come to a final equilibrium temperature. Part A Which of the following statements is correct concerning the temperature changes of both substances? (Ignore the signs of the temperature changes in your answer.) Which of the following statements is correct concerning the temperature changes of both substances? (Ignore the signs of the temperature changes in your answer.) The temperature change of the copper is equal to the temperature change of the water. The temperature change of the water is greater than the temperature change of the copper. The temperature change of the copper is greater than the temperature change of the water.
Physics
1 answer:
Valentin [98]3 years ago
7 0

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

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Pure water, instead, is a very bad conductor, this means its resistivity is very high, of order of k \Omega \cdot m (10^3 \Omega m ). Even without knowing the precise value of the pure water resistivity, we can estimate the ratio between the pure water resistivity and the silver resistivity by comparing the two orders of magnitude:
r= \frac{10^3 \Omega m}{10^{-8} \Omega m}  \sim 10^{11}

Therefore, we can say that the correct answer is
3 \cdot 10^{11} : 1
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A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally
Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, g=-9.8 m/s^2, downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

v_x = +2.60 m/s

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

v_y = u_y +gt

where

u_y = 0 is the initial vertical velocity, which is zero

g=-9.8 m/s^2 is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

v_y = 0+(-9.8)(1.75)=-17.2 m/s

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0
3 years ago
A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of
Leto [7]

Answer: 2.27\ m/s^2

Explanation:

Given

Length of the race track L=200\ m

the radius of curvature of the track r=29.5\ m

time taken to run on track is t=24.4\ s

Speed of runner is

\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s

Centripetal acceleration is

\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2

5 0
3 years ago
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