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3 years ago

Can someone please solve this

Physics

1 answer:

Zina [86]3 years ago

8 0

Answer:

The gravity is pulling the diver downwards but the rotation of the body means gravity cant pull him down as quickly

Explanation:

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A heat engine has three quarters the thermal efficiency of a carnot engine operating between temperatures of 65Â°C and 435Â°C if

const2013 [10]

Answer:

The value is $P_e = 31275.2 \ W$

Explanation:

From the question we are told that

The efficiency of the carnot engine is $\eta$

The efficiency of a heat engine is $k = \frac{3}{4} * \eta$

The operating temperatures of the carnot engine is $T_1 = 65 ^oC =338 \ K$ to $T_2 = 435 ^oC = 708 \ K$

The rate at which the heat engine absorbs energy is $P = 44.0 kW = 44.0 *10^{3} \ W$

Generally the efficiency of the carnot engine is mathematically represented as

$\eta = [ 1 - \frac{T_1 }{T_2} ]$

=> $\eta = [ \frac{T_2 - T_1}{T_2} ]$

=> $\eta = 0.3856$

Generally the efficiency of the heat engine is

$k = \frac{3}{4} * 0.3856$

=> $k = 0.2892$

Generally the efficiency of the heat engine is also mathematically represented as

$k = \frac{W}{P}$

Here W is the work done which is mathematically represented as

$W = P - P_e$

Here $P_e$ is the heat exhausted

$k = \frac{P - P_e}{P}$

=> $0.2892 = \frac{44*10^{3} - P_e}{44*10^{3}}$

=> $P_e = 31275.2 \ W$

8 0

3 years ago

How does an accretion disk around a neutron star differ from an accretion disk around a white dwarf?.

user100 [1]

Well the accretion disk around a neutron star is much hotter and emits higher energy radiation

4 0

2 years ago

The _____ of an object consists of its speed and direction.

ELEN [110]

Velocity would be an answer

3 0

2 years ago

Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it

d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s, 1) v = $\sqrt{0.55^2 + 19.6 y}$ , 2) v = 2.05 m / s,

3) d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

Q = v A

where Q is the flow rate, A is area and v is the velocity

the area of a circle is

A = π r²

radius and diameter are related

r = d / 2

substituting

A = π d²/4

Q = π/4 v d²

let's reduce the magnitudes

v = 0.55 m / s = 55 cm / s

let's calculate

Q = π/4 55 1.96²

Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

v = $\sqrt{0.55^2 + 2 \ 9.8\ y}$

v = $\sqrt{0.55^2 + 19.6 y}$

2) ask to calculate the velocity for y = 0.2 m

v = $\sqrt{0.55^2 + 19.6 \ 0.2}$

v = 2.05 m / s

3) We write the continuous equation for this point 2

Q = v₂ A₂

A₂ = Q / v₂

let us reduce to the same units of the SI system

Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

A₂ = 165.95 10⁻⁶ / 2.05

A₂ = 80,759 10⁻⁶ m²

area is

A₂ = π/4 d₂²

d₂ = $\sqrt{4 A_2 / \pi }$

d₂ = $\sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }$

d₂ = 10.14 10⁻³ m

d₂ = 1.014 cm

4 0

3 years ago

An electric heating element has a resistance of 16 Ω and is connected to a voltage of 120 V. How much current will flow in this

sergeinik [125]

This is weird.

All three 'choices' are true.

Line um up. (a) shows how to solve the problem. (b) does it. and (c) is the answer.

7 0

3 years ago