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NikAS [45]
2 years ago
12

At what height above the surface of the earth does the weight of an object decrease to 0.8​

Engineering
1 answer:
prohojiy [21]2 years ago
8 0

Answer: That'd be about 6400km upwards, i belive.

Explanation:

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A safety interlock module operates by monitoring the voltage from the
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Answer: its an Ignition coil

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A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is 20.3 kJ/K
Mashcka [7]

Answer:

W= 8120 KJ

Explanation:

Given that

Process is isothermal ,it means that temperature of the gas will remain constant.

T₁=T₂ = 400 K

The change in the entropy given ΔS = 20.3 KJ/K

Lets take heat transfer is Q ,then entropy change can be written as

\Delta S=\dfrac{Q}{T}

Now by putting the values

20.3=\dfrac{Q}{400}

Q= 20.3 x 400 KJ

Q= 8120 KJ

The heat transfer ,Q= 8120 KJ

From first law of thermodynamics

Q = ΔU + W

ΔU =Change in the internal energy ,W=Work

Q=Heat transfer

For ideal gas ΔU  = m Cv ΔT]

At constant temperature process ,ΔT= 0

That is why ΔU  = 0

Q = ΔU + W

Q = 0+ W

Q=W= 8120 KJ

Work ,W= 8120 KJ

8 0
3 years ago
1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by
zvonat [6]

Answer:

299.36 feet

Explanation:

To \ find  \   the  \ distance \  of  \ the  \ ball \  from  \ the \ home  \ plate.  \\ \\ From  \ the  \ given  \ information:

Height \ h = 4 \ ft

Initial \ speed \ V_o = 98 \ ft/s ec

The  \ angle \  \theta = 45^0

Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s

U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}

U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}

So;

S_y = u_y t - \dfrac{1}{2}gt^2

-1 =\dfrac{98}{\sqrt{2}}t - \dfrac{1}{2}*32*1.85t^2

By solving:

t_1 = 4.32 \ sec

Thus;

horizontal \ distance = U_x t

= \dfrac{98}{\sqrt{2}}\times 4.32

\mathbf{=299.36  \ feet}

\mathbf{Thus \ , the  \  distance \ from \  the  \ home  \ plate \  =  \ 299.36  \ feet}

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3 years ago
What should you consider when choosing the type of head protection
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Material,Flexibility,Comfort,Workplace Hazards

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