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krok68 [10]
3 years ago
5

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

reservoir. during that time, what is the maximum amount of work wmax that the engine might have performed?
Physics
1 answer:
tino4ka555 [31]3 years ago
6 0

q_{c} = Heat released to cold reservoir

q_{h} = Heat released to hot reservoir

W_{max} = maximum amount of work

t_{c} = temperature of cold reservoir

t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



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Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound
Sophie [7]

Answer:

a. Wavelength = λ = 20 cm

b. Next distance of maximum intensity will be 40 cm

Explanation:

a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.

Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.

This we get the equation:

(nλ + λ/2) - nλ = 30-20

λ/2 = 10

λ = 20 cm

b. at what distance, sound intensity will be maximum again.

For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.

= 30 + λ/2

= 30 + 20/2

=30+10

=40 cm

7 0
3 years ago
A supersonic jet, with a mass of 21,000 kg, departs from its home airbase with a velocity of 400 m/s due east. What is the jet’s
disa [49]

Answer:

Momentum  P is 840000kgm/s or 8.4 × 10^6

Explanation:

Data :

        Mass = 21000 kg

        Velocity = 400 m/s

So momentum is given as

      P = mv

      P = 21000×400

    P = 8400000 kgm/s

       P = 8.4 × 10^6

6 0
2 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
3 years ago
Help me. It's question 2 I need the answer for
Alex73 [517]
Ok, let me see if I can help

Sound is caused by vibrations. These can pass through a solid, liquid, and gas. But not through vacuum because there are no particles
6 0
3 years ago
An airplane starts from rest and accelerates at a constant rate of 3.00 m/s2 for 30.0 s before leaving the ground. a. How far di
rosijanka [135]

The formula we can use in this case is:

d = v0t + 0.5 at^2

v = at + v0

where,

d = distance travelled

v0 = initial velocity = 0 since at rest

t = time travelled

a = acceleration

v = final velocity when it took off

 

a. d = 0 + 0.5 * 3 * 30^2

d = 1350 m

 

b. v = 3 * 30 + 0

<span>v = 90 m/s</span>

8 0
3 years ago
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