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2 years ago

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Lee pushes horizontally with a force of 75 n on a 36 kg mass for 10 m across a floor. calculate the amount of work lee did. answ

er in units of j.

2 answers:

svlad2 [7]2 years ago

8 0

To find work, you use the equation: W = Force X Distance X Cos (0 degrees)
Following the Law of Conservation of Energy, energy cannot be destroyed nor created.

So you would do 75 N x 10m x Cos (0 degrees)= 750 J

Marysya12 [62]2 years ago

7 0

Answer:

The work done by Lee is 3528 Joules.

Explanation:

Given that,

Mass of the object, m = 36 kg

Force applied by Lee on the object, F = 75 N

The object moves to a distance, d = 10 m

We need to find the amount of work done by Lee. It is given by the combination of force and displacement of an object. It is given by :

$W=F{\cdot} d$

$W=mg{\cdot} d$

$W=36\times 9.8\times 10$

W = 3528 Joules

So, the work done by Lee is 3528 Joules. Hence, this is the required solution.

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In a closed system, what happens to the total energy of the system as energy conversion takes place?

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Answer:

In a closed system, the total energy is conserved or remains the same as energy transformations take place.

Explanation:

The law of conservation of energy states that energy cannot be created or destroyed but can be transformed from one form to another.

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2 years ago

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2 years ago

Suppose you want to use this human engine to lift a 2.35 kg box from the floor to a tabletop 1.30 m above the floor. How much mu

morpeh [17]

Answer:

The increase in the gravitational potential energy is 29.93 joules.

Explanation:

Given that,

Mass of the box, m = 2.35 kg

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We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :

$U=mgh$

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So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.

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2 years ago

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

So

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

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2 years ago