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2 years ago

Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal

Physics

2 answers:

Serhud [2]2 years ago

6 0

Mass doesn't matter on amount of work done .We can calculate amount of work done through Force and Displacement

Force=87N
Displacement=4.5m

$\boxed{\sf W=Fd}$

W denotes to work done
F denotes to force
d denotes to displacement

$\\ \sf\longmapsto Work\:Done=87(4.5)$

$\\ \sf\longmapsto Work\:done=391.5J$

elena-s [515]2 years ago

5 0

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

W = F║d
where the force is parallel to the displacement

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

W = (87 N)(4.5 m)
W = 391.5 J

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

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fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

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3 years ago

A ball falls from the top of a building. As it falls, its speed increases. Which type of energy is the ball gaining as it falls?

Elina [12.6K]

<h2>Hello!</h2>

The answer is: B. Kinetic energy

Since the ball is falling, speed increases because the gravity acceleration is acting. When speed increases, the kinetic energy increases too, so the ball is gaining kinetic energy.

The gravity acceleration is equal to $9.81\frac{m}{s^{2}}$ , it means that when falling, the ball will increase it's speed 9.81m every second.