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2 years ago

An adiabatic closed system is accelerated from 0 m/s to 30 m/s. Determine the specific energy change of this system, in kJ/kg.

Physics

1 answer:

snow_tiger [21]2 years ago

6 0

Answer:

Explanation:

initial velocity, u = 0 m/s

final velocity, v = 30 m/s

Let the mass is m .

Initial kinetic energy, Ki = 1/2 mu² = 0

Final kinetic energy, Kf = 1/2 mv² = 0.5 m x 30 x 30 = 450 m

Specific energy change = change in kinetic energy per unit mass

= 450 m / m = 450 J/kg = 0.45 kJ/kg

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Two convex thin lenses with focal lengths 10.0 cm and 20.0 cm are aligned on a common axis, running left to right, the 10-cm len

love history [14]

Answer:

Explanation:

f1 = 10 cm

f2 = 20 cm

u = Object distance = 15 cm

Distance between lenses = 20 cm

For first lens image distance

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}\\\Rightarrow \frac{1}{v}=\frac{1}{30}\\\Rightarrow v=30\ cm$

Distance from second lens is 10 cm to the right

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{-10}\\\Rightarrow \frac{1}{v}=\frac{3}{20}\\\Rightarrow v=6.67\ cm$

The final image will appear as +6.67 cm

3 0

2 years ago

The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe

Novosadov [1.4K]

Answer:

a) a= 8.33 m/s², T = 12.495 N , b) a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

fr₁ + fr₂ = ma

The equation for the force of friction is

fr = μ N

Y Axis

N - W₁ –W₂ = 0

N = W₁ + W₂

N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

N = 2 m g

We replace

μ₁ 2mg + μ₂ mg = m a

a = (2μ₁ + μ₂) g

a = (2 0.30 + 0.25) 9.8

a= 8.33 m/s²

Let's look for cable tension with beam 2

T = m₂ a

T = 1500 8.33

T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

fr = m₂ a₂

N- w₂ = 0

N = W₂ = mg

μ₂ mg = m a₂

a = μ₂ g

a = 0.25 9.8

a = 2.45 m / s²

4 0

2 years ago

Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD

NeX [460]

Answer:

KE= 1/2mv²

Explanation:

The kinetic energy of a body is the energy possessed by virtue of the body in motion

Given the parameters

m which is the mass of the body

v which is the velocity of the body too

K.E = kinetic energy

The expression for the kinetic energy of a body is given as

KE= 1/2mv²

3 0

2 years ago

The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of

Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0

2 years ago

luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that

Andreas93 [3]

Answer:

a) W = 643.5 J, b) W = -427.4 J

Explanation:

a) Work is defined by

W = F. x = F x cos θ

in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero

W = F x

let's calculate

W = 165 3.9

W = 643.5 J

b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes

sin θ = Wₓ / W

cos θ = Wy / W

Wₓ = W sinθ = mg sin θ

Wy = W cos θ

the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero

W = Wₓ x cos 180

W = - mg sin 34 x

let's calculate

W = -20 9.8 sin 34 3.9

W = -427.4 J

The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.

3 0

2 years ago