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2 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Physics

1 answer:

Likurg_2 [28]2 years ago

8 0

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

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The force applied by the man is 60 N

Explanation:

We can solve this problem by applying Newton's second law, which states that:

$\sum F = ma$ (1)

where

$\sum F$ is the net force acting on the child+cart

m is the mass of the child+cart system

a is their acceleration

In this problem, we have:

m = 30.0 kg is the mass

$a=1.50 m/s^2$

And there are two forces acting on the child+cart system:

The forward force of pushing, F
The force resisting the cart motion, R = 15.0 N

Therefore we can write the net force as

$\sum F = F -R$

where R is negative since its direction is opposite to the motion

So eq.(1) can be rewritten as

$F-R=ma$

And solving for F,

$F=ma+R=(30.0)(1.50)+15.0=60 N$

Learn more about Newton's second law:

brainly.com/question/3820012

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2 years ago

Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo

LenaWriter [7]

Answer:

$6.33\times 10^8\ kg\cdot m/s$

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = $33\times \frac{5}{18}=9.17\ m/s$

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

= $(6.9\times 10^7\ kg)(9.17\ m/s)$

= $6.33\times 10^8\ kg\cdot m/s$

Therefore, the magnitude of ship's momentum is $6.33\times 10^8\ kg\cdot m/s$ .

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2 years ago

How much power does it take to lift a 24 kg gift box 6m above the floor in 4 s?

Mrac [35]

Answer:

Explanation:

Step one:

given

mass m= 24kg

distance moved= 6m

time taken= 4seconds

Step two:

Required

power

but work done is the force applied at a distance, and the power is the work done time the time taken

Work done= F*D

F=mg

W= mg*D

W=24*9.81*6

W=1412.6J

Power P= work * time

P=1412.6*4

p=5650.5W

P=5.6kW

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2 years ago

After soccer practice, Coach Miller goes to the roof of the school to retrieve the errant soccer balls. The height of the school

blsea [12.9K]

With gravitational acceleration at 9.8, initial height at 3.5m and distance at 22m the initial horizontal velocity is 26.03 ms and the flight time is .845 seconds

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2 years ago

Consider the reaction data. A ⟶ products T ( K ) k ( s − 1 ) 225 0.385 525 0.635 What two points should be plotted to graphicall

lutik1710 [3]

Answer:

Plot ln K vs 1/T

(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol

Explanation:

This is an example of the Arrhenius equation:

$k = Ae^{-E_{a}/RT}\\\text{Take the ln of each side}\\\ln k = \ln A - \dfrac{E_{a}}{RT}\\\\\text{We can rearrange this to give}\\\ln k = - \dfrac{E_{a}}{R}\dfrac{1}{T} + \ln A\\\\y = mx + b$

Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA

Data:

$\begin{array}{cccc}\textbf{k/s}\mathbf{^{-1}} &\mathbf{\ln k} & \textbf{T/K} & \mathbf{1/T(K^{-1})}\\0.285 & -0.9545 & 225 &0.004444\\0.635 & -0.4541 & 525 & 0.001905\\\end{array}$

Calculations:

(a) Rise

Δy = y₂ - y₁ = -0.9545 - (-0.4541) = -0.9545 + 0.4541 = -0.5004

(b) Run

Δx = x₂ - x₁ = 0.004 444 - 0.001 905 = 0.002 539 K⁻¹

(c) Slope

Δy/Δx = -0.5004/0.002 539 K⁻¹ = -197.1 K⁻¹

(d) Activation energy

Slope = -Eₐ/R

Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol

4 0

2 years ago