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3 years ago

The force of gravity down on the boy= 300N

Physics

2 answers:

Leviafan [203]3 years ago

8 0

The net force on the boy is zero.

Based on this analysis, I would expect the boy to <u>not accelerate</u>.

kompoz [17]3 years ago

3 0

The net force of the boy is 300N=><=300N which equals a net force of 0N.

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Artemon [7]

Answer:

Both speed and velocity are changing.

Explanation:

They are both going up so both are changing

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3 years ago

Develop an equation (with a proportionality constant) that describes the relationship between the gravitational force (fgrav), t

ivolga24 [154]

According to newton's law of gravitation, the gravitational force(F) is directly proportional to the product mass of the moon(Mm) and the mass of the planet (Mp) and it is inversely proportional to the square of the separation between them.

Fg ∝ (Mp)(Mm) →(1)

Fg ∝ 1/d²→(2)

Combining equation (1) and (2),

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This is an equation that describes the relation between mass of moon (Mm) and mass of planet (Mp) and separation(d) between them.

To support the claim in favuor of this equation we use this equation to obtain the value of acceleration due to gravity on earth.

Let m be the mass of an object on earth then Fg between earth (Mp) and mass of an object is obtained by:

Fg = G(Mp)(m)/R², where R= Radius of earth

This force is equal to the weight of an object i.e.,

g= G(Mp)/R²

Putting the values of G, Mp and R , we get, g=9.81 m/s²

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1 year ago

How much heat is required to convert 2.55g of water at 28 degrees c to steam?

pantera1 [17]

There are two different processes here:
1) we must add heat in order to bring the temperature of the water from $28^{\circ}C$ to $100^{\circ}C$ (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
$Q_1=m C_s \Delta T$
where
$m=2.55 g$ is the water mass
$C_s = 4.18 g/J^{\circ}C$ is the water specific heat
$\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C$ is the variation of temperature of the water
If we plug the numbers into the equation, we find
$Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J$

2) The heat needed for process 2) is
$Q=m L_e$
where
$m=2.55 g$ is the water mass
$L_e = 2264.7 J/g$ is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
$Q_2=(2.55 g)(2264.7 J/g)=5775.0 J$

So, the total heat needed for the whole process is
$Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J$

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