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meriva
1 year ago
7

A piston–cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first expanded isothermally to 500 kPa

, then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the boundary work for each process and the net work of the cycle.
Physics
1 answer:
Paraphin [41]1 year ago
3 0

Answer:

<h2>jeusYgwyhedswusjsj</h2>

Explanation:

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Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent
Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

d = 2\pi r

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

v = \frac{2\pi r}{T}

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

6 0
2 years ago
Read 2 more answers
A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,
Annette [7]

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

d_x = 10 + 2= 12 blocks

Total displacement towards South

d_y = 5 block

now to find the net displacement we can use vector addition

d = \sqrt{d_x^2 + d_y^2}

d = \sqrt{12^2 + 5^2}

d = 13 blocks

<em>so magnitude of net displacement will be equal to 13 blocks</em>

6 0
1 year ago
A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a fri
liubo4ka [24]

Answer:

T = 7.64 kN

F_y = 0.52 kN(Downwards)

F_x = 3.23 kN (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

F(4.50 cos\theta) + mg(2cos\theta) = T \times 3

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T

T = 7.64 kN

Now force balance in vertical direction

F + mg = Tsin65 + F_y

5 + 1.40 = 7.64 sin65 + F_y

F_y = 0.52 kN(Downwards)

Force balance in horizontal direction

F_x = T cos65

F_x = 7.64 cos65

F_x = 3.23 kN (Towards Left)

7 0
2 years ago
8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
A sleeping 68 kg man has a metabolic power of 79 w .
Lesechka [4]
 <span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>
3 0
2 years ago
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