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3 years ago

The can-chiller decreases the temperature of the liquid in the can by 15 °C.

Physics

1 answer:

LUCKY_DIMON [66]3 years ago

7 0

Answer:

Energy transferred, Q = 20790 J

Explanation:

Given that,

The change in temperature of the liquid, $\Delta T = 15^{\circ} C$

The mass of liquid, m = 0.33 kg

The specific heat capacity of the liquid is 4200 J/kg°C.

We need to find the energy transferred from the liquid as it cools. The energy transferred is given by :

$Q=mc\Delta T\\\\Q=0.33\times 4200 \times 15\\\\Q=20790\ J$

So, 20790 J of energy is transferred from the liquid as it cools.

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Can someone help me?

Cloud [144]

What do we know that might help here ?

-- Temperature of a gas is actually the average kinetic energy of its molecules.

-- When something moves faster, its kinetic energy increases.

Knowing just these little factoids, we realize that as a gas gets hotter, the average speed of its molecules increases.

That's exactly what Graph #1 shows.

How about the other graphs ?

-- Graph #3 says that as the temperature goes up, the molecules' speed DEcreases. That can't be right.

-- Graph #4 says that as the temperature goes up, the molecules' speed doesn't change at all. That can't be right.

-- Graph #2 says that after the gas reaches some temperature and you heat it hotter than that, the speed of the molecules starts going DOWN. That can't be right.

4 0

3 years ago

Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two op

Nostrana [21]

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

Where:

$v_{ex}$ is relationship between exhaust velocity and specific impulse

$\frac{m_{f}}{m_{e}}$ is the porpellant fraction, also written as MR.

The relationship $v_{ex}$ is: $v_{ex} = g_{0}.Isp$

To determine the fraction:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

$ln(MR) = \frac{v}{v_{ex}}$

Knowing that change in velocity is Δv = 9.6km/s and $g_{0}$ = 9.81m/s²

Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

Part 1: Isp = 450s

$ln(MR) = \frac{v}{v_{ex}}$

ln(MR) = $\frac{9.6.10^{3}}{9.81.450}$

ln (MR) = 2.17

MR = $e^{2.17}$

MR = 8.76

Part 2: Isp = 2000s

$ln(MR) = \frac{v}{v_{ex}}$

ln (MR) = $\frac{9.6.10^{3}}{9.81.2.10^{3}}$

ln (MR) = 0.49

MR = $e^{0.49}$

MR = 1.63

8 0

3 years ago

Question 6

Kisachek [45]

6. gas

7. liquid

8. gas

9. does not change

10. boiling point

7 0

3 years ago

Two 0.55-kg basketballs, each with a radius of 19 cm , are just touching. You may want to review (Pages 399 - 401) . Part A How

Shtirlitz [24]

Answer:

ΔU = - 9,179 10-11 J

Explanation:

For this exercise the two basketballs are linked by gravitational interaction, so we can use gravitational energy

U = - G m₁m₂ / R

In this case the mass is equal and the initial distance is r₁ = 19 cm = 0.19 m

U₁ = - G m² / r₁

let's calculate

U₁ = - 6.67 10⁻¹¹ 0.55² / 0.19

U₁ = - 10,619 10⁻¹¹ J

when its centers are separated it is at r₂ = 1.4 m

U₂ = - 6.67 10⁻¹¹ 0.55² / 1.4

U₂ = - 1.44 10-11 J

the energy between these two points is

ΔU = U₂ - U₁

ΔU = (-1.44 +10.619) 10-11

ΔU = - 9,179 10-11 J

7 0

4 years ago

Step 2: Apply NEwton's second law Apply ∑Fy = may , what should ay be equal to, since the block doesn't move in the y direction

andrey2020 [161]

Answer:

∑Fy = 0, because there is no movement, N = m*g*cos (omega)

Explanation:

We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.

If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.

3 0

3 years ago