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4 years ago

what happens to the gravitational force when the distance between them stays the same and the mass of both objects is doubled

Physics

1 answer:

Dimas [21]4 years ago

7 0

Answer:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.

Explanation:

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When using a different calorimeter, and mixing 50 ml of hot water at 65 degrees c with 60 ml of water in the calorimete

Paha777 [63]

The specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.

<h3>Conseervation of energy</h3>

The heat capacity of the calirometer is determined by applying the principle of conservation of energy.

Heat lost by the hot water = Heat gained by the calirometer

$Q _w = Q_c\\\\M_w C_w\Delta \theta _w = M_c C_c\Delta \theta _c$

where;

M is mass

mass = density x volume = ρV

Density of water = 1 g/ml

Mass of hot water = 1 x (50) = 50 g

Mass of water in calorimeter = 1 x (60) = 60 g

<h3>Equilibrium temperature</h3>

$\Delta T_c = 5.5\\\\T - 25 = 5.5\\\\T = 30.5 \ ^0C$

<h3>Specific heat capacity of the calirometer</h3>

$50 \times 4.184 \times (65 - 30.5) = 60 \times C_c \times (30.5 - 25)\\\\7217.4 = 330C_c\\\\C_c = \frac{7217.4}{330} \\\\C_c = 21.87 \ J/g^0C$

Thus, the specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.

The complete question is below

When using a different calorimeter, and mixing 50 ml of hot water at

65 degrees C with 60 ml of water in the calorimeter at 25 degrees C, the temperature of the calorimeter increased by 5.5 degrees C.

a. Calculate the heat capacity of this calorimeter?

Learn more about heat capacity here: brainly.com/question/16559442

5 0

2 years ago

A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5

koban [17]

Answer:

V = 38.0 L

Explanation:

As we know that number of moles will remains conserved inside the balloon

so we will have

$moles = \frac{PV}{RT}$

here we have

$\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}$

now we have

$P_1 = 760 mm Hg$

$P_2 = 76 mm Hg$

$V_1 = 5.00 L$

$T_1 = 20^o C = 293 K$

$T_2 = -50^o C = 223 K$

$\frac{(760mm Hg)(5L)}{R(293)} = \frac{(76mm Hg)(V)}{R(223)}$

$V = 38.0 L$

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Answer:

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Explanation:

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3 years ago

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