Ask question

Ask question

All categories

aleksandrvk [35]

2 years ago

5

0.2000 kg of water at 20.00°C is contained in a 0.1000-kg copper container. The container is shaken vigorously for 10.00 minutes

to cause the temperature to rise to 22.00°C. Calculate the work done on the system and the heat supplied to the system. The specific heat of copper is 385.0 J/kg and of water is 4200.0 J/kg

1 answer:

Tresset [83]2 years ago

8 0

Work done =1757JHeat supplied=0

You might be interested in

What happens to a mirror when a ray is directed at it

mixas84 [53]

Nothing happens to the mirror.

However, if the ray is within some suitable range of
wavelengths, the ray is reflected from the mirror's surface.

3 0

2 years ago

A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits

Nikitich [7]

Answer:

v=77.62 m/s

Explanation:

Given that

h= - 300 m

speed of the bird ,u= 5 m/s

Lets take Speed of the berry when it hit the ground = v m/s

we know that ,if object is moving upward

v² = u² - 2 g h

u=Initial speed

v=Final speed

h=Height

Now by putting the values

v² = u² - 2 g h

v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)

v² =25 + 20 x 300

v² ==25 + 6000

v² =6025

v=77.62 m/s

Therefore the final speed of the berry will be 77.62 m/s.

5 0

3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

2 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

3 years ago

12) Water flows through a horizontal pipe of cross-sectional area 10.0 cm2 at a pressure of 0.250 atm with a flow rate is 1.00 L

masha68 [24]

Answer:

The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]

Explanation:

We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.

For this case, we don't use the elevation difference and therefore those terms can be cancelled.

When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.

5 0

3 years ago