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2 years ago

7

A force of 77 N is used to move a desk 1.5 m. How much work was done?

1 answer:

Sloan [31]2 years ago

3 0

Answer:

As we know,

work = 77 N × 1.5m

therefore 115.5 J work was done

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Why does the fed rarely change the reserve requirement

Brut [27]

The answer is:

It can be disruptive to the whole banking system.

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3 years ago

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A laboratory experiment produces a double-slit interference pattern on a screen. The point on the screen marked with a white dot

Alenkasestr [34]

Explanation:

Diffraction is a characteristic phenomenon for waves, and is based on their variance to find an obstacle or a crack. The diffraction and interference are inseparable phenomena, which are so harder to identify. Since diffraction is a specific type of interference.

if the point are in the order 2, and

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3 years ago

t requires1200 kg of coal to produce the energy needed to make 1.0 kg of aluminum metal. If a single soda can requires approxima

Lesechka [4]

Answer:

The mass of coal is 108 kg.

Explanation:

Given that,

Energy of coal = 1200 kg

Mass of aluminum = 1.0 kg

Energy required for single soda can = 15 g of Al

Energy required for 6 pack of cans = $6\times15=90\ g\ of\ Al$

We need to calculate the mass of coal

Using formula of mass

$\text{mass of coal}=\dfrac{\text{Energy of coal}\times\text{Energy required for 6 pack of cans}}{\text{Mass of aluminum}}$

Put the value into the formula

$m=\dfrac{1200\times90\times10^{-3}}{1.0}$

Put the value into the formula

$m=108\ Kg$

Hence, The mass of coal is 108 kg.

6 0

3 years ago

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi

trapecia [35]

Answer:

a) $F_p=882N$

b) $P=4410W$

c) $V_p'=24135$ , $n=15.2\%$

Explanation:

From the question we are told that:

Mass $M=1500kg$

Velocity $v=4.9m/s$

Coefficient of Rolling Friction $\mu=0.06$

a)

Generally the equation for The Propulsion Force is mathematically given by

$F_p=\mu*mg$

$F_p=0.06*1500*9.81$

$F_p=882N$

b)

Therefore Power Required at

$V_p=5.0m/s$

$P=F_p*V_p$

$P=882*5$

$P=4410W$

c)

$V_p' =15mpg$

$V_p'=15*\frac{1609}$

$V_p'=24135$

Generally the equation for Work-done is mathematically given by

$W=F_p*V_p'$

$W=882*15*1609$

$W=2.13*10^7$

Therefore

Efficiency

$n=\frac{W}{E}*100\%$

Since

Energy in one gallon of gas is

$E=1.4*10^8J$

Therefore

$n=\frac{2.1*10^7}{1.4*10^8}*100\%$

$n=15.2\%$

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3 years ago

It was recorded that the temperature of a body was 320 degree F determine the value of the temperature in kelvin

Thepotemich [5.8K]

Answer:

433.15K

Explanation:

(320°F − 32) × 5/9 + 273.15 = 433.15K

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3 years ago