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kotykmax [81]
3 years ago
9

The gravitational force exerted on a solid object is 5.00N. When the object is suspended from a spring scale and submerged compl

etely in water the scale reads 3.50N. Find the density of the object.
Physics
1 answer:
marin [14]3 years ago
4 0

Answer:

3333.33 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to its volume.

The unit of density is kg/m³.

From Archimedes principle,

R.d = W/U = D/D'

Where R.d = relative density, W = weight of the object in air, u = upthrust in water, D = Density of the object, D' = Density of water.

W/U = D/D'

making D the subject of the equation

D = D'(W/U).......................... Equation 1

Given: W = 5.0 N, U = 5.0 -3.5 = 1.5 N, D' = 1000 kg/m³

Note: U = lost in weight = weight in air - weight in water

Substitute into equation 1

D = 1000(5/1.5)

D = 3333.33 kg/m³

Thus the density of the object = 3333.33 kg/m³

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Q is the heat absorbed by the engine to do the work

In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
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5g of ammonium nitrate was dissolved in 60g of water in an insulated container. The temperature at the start of the reaction was
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Energy absorbed by the reaction or energy lost by the water to the reaction,Q.

Mass of the the reaction  ,m = 60 g

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Change is temperature=\Delta T=19^oC-23^oC=-4^oC

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Negative sigh indicates that energy was given by the water to the reaction.

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A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
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Answer:

451.13 J/kg.°C

Explanation:

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Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

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If a roller coaster car had 40,000 J of gravitational potential energy when at rest on the top of a hill how much kinetic energy
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Answer:

K.E = 30,000 J

Explanation:

Given,

The potential energy of the roller coaster car, P.E = 40000 J

The kinetic energy at height h/4, K.E = ?

According to the law of conservation of energy, the total energy of the system is conserved.

At height 'h', the total energy is,

                                    P.E = mgh

                                     K.E = 0

At height 'h/4', the total energy is

                                     P.E + K.E = mgh

                                     P.E = mgh/4

                                     K.E = 1/2 mv²

Therefore,

                                   mgh/4 + 1/2 mv² = mgh

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Hence,

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Substituting in the K.E equation

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                                         = 30000 J

Hence, the K.E of the roller coaster car is, K.E = 30000 J

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