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3 years ago

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A ball is thrown horizontally at 6.0 m/s from a cliff 80 meters high. How far from the base of the cliff will the ball hit the g

round?

1 answer:

UNO [17]3 years ago

6 0

Answer:

160 meters I think

Explanation:

if I'm wrong I'm a troop juSt trying to get points to ask a couple of questions

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What is carnot engine.express the relation for its efficiency?

patriot [66]

Answer:

Explanation:

The Carnot cycle is a special case of a thermodynamic cycle that produces an ideal gas and consists of two isothermal processes and two adiabatic processes. This cycle is a theoretical solution given by Sadi Karnot to refine heat engines for their efficient use.

The formula for the coefficient of efficiency is:

η = (Q₁ - Q₂) / Q₁ = (T₁ - T₂) / T₁

Where Q₁ is is the amount of heat of the heater supplied to the working body and Q₂ is the amount of heat that the working body transfers to the refrigerator according to this T₁ is the temperature of the heater T₂ is the temperature of the refrigerator.

This formula provides a theoretical limit for the maximum value of the coefficient of efficiency of heat engines.

God is with you!!!

6 0

3 years ago

Read 2 more answers

Consider that 168.0 J of work is done on a system and 305.6 J of heat is extracted from the system. In the sense of the first la

Ilia_Sergeevich [38]

To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

$dQ = dU + dW$

Where,

Q = Heat

U = Internal Energy

By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore

$W = -168J \righarrow$ Work is done ON the system

$Q = -305.6J \rightarrow$ Heat is extracted FROM the system

Therefore the value of the Work done on the system is -158.0J

3 0

3 years ago

Tarzan swings on a 31.0 m long vine initially inclined at an angle of 36.0◦ with the vertical. The acceleration of gravity if 9.

Gennadij [26K]

Answer:

$v=10.777m/s$

Explanation:

Tarzan swing can be thought of as change in potential energy by going from higher location We solve for height of beginning of the swing by using simple cosine equation:

So

$31Cos36=25.08\\E_{potential}=mgh\\$

ΔE=mg(h₂-h₁)

$=m*9.81(31-25.08)\\$

The potential energy of Tarzan initial position is converted into kinetic energy of his swing.By using kinetic equation

$E_{kinectic}=P_{potential}\\1/2mv^{2}=m*9.81(31.0-25.08)\\(1/2)v^{2}=9.81(31.0-25.08)\\0.5v^{2}=58.07\\v^{2}=58.07/0.5\\v=\sqrt{58.07/0.5}\\ v=10.777m/s$

8 0

3 years ago

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

<u>From the Fourier's law of conduction:</u>

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&<u>For Styrofoam-plywood interface from inside:</u>

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

4 0

3 years ago

Explain why nuclear fission and nuclear fusion release large amounts of energy

levacccp [35]

Answer:

Because of the formula $E=mc^2$

Explanation:

In this problem we are describing two different processes:

Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei

Nuclear fusion occurs when two (or more) light nuclei fuse together producing a heavier nucleus

In both cases, the total mass of the final products is smaller than the total mass of the initial nuclei.

According to Einsten's formula, this mass difference has been converted into energy, as follows:

$E=\Delta mc^2$

where:

E is the energy released in the reaction

$\Delta m$ is the mass defect, the difference between the final total mass and the initial total mass

$c=3.0 \cdot 10^8 m/s$ is the speed of light

From the formula, we see that the factor $c^2$ is a very large number, therefore even if the mass defect $\Delta m$ is very small, nuclear fusion and nuclear fission release huge amounts of energy.

8 0

3 years ago