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ella [17]
3 years ago
13

A 0.65 kg rock is projected from the edge of

Physics
1 answer:
lapo4ka [179]3 years ago
8 0

Answer:

180.45m

Explanation:

Given that

mass m= 0.65kg

let H be height of building

angle of projection =40°

initial velocity  v up = 10.5 sin 40 = 6.75 m/s

horizontal velocity u= 10.5 cos 40 = 8.04 m/s

horizontal distance x = 19.6

x=ut

x= u t =

x= 8.04 t

t=19.6/8.04  = 2.44 seconds

applying the formula

h = H + Vt -1/2gt^2

0 = H + 6.75 * 2.44 - 4.9 * 2.44^2

solving for H we have

0=H + 6.75 * 2.44 - 4.9 * 5.9536

0=H + 6.75 * 2.44 - 29.17264

0=H + 6.75 *-26.73264

0=H -180.44532

H=180.45m

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ASHA 777 [7]

Answer:

  • +1.8W
  • 2,160J

Explanation:

First, we need to solve for the power absorbed by the bulb.

Formula [ p = vi ]

p = 12 * 150 * 10^-3

p = +1.8W

Second, we can solve for the power aborbed in a interval of 20 minutes.

Formula [ e = vit ]

e = 12 * 150 * 10^-3 * 20 * 60

e = 2,160J

Just a note that "20 * 60" represents the time given in seconds.

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7 0
2 years ago
A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and
Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

W=mg

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

\rho_a = 1.29 kg/m^3 is the air density

V=329 m^3 is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N

(c) 1524 N

The mass of the helium gas inside the balloon is

m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg

where \rho_h is the helium density; so we the total mass of the balloon+helium gas inside is

m=m_h+m_b=59 kg+210 kg=269 kg

So now we can find the weight of the balloon:

W=mg=(269 kg)(9.8 m/s^2)=2635 N

And so, the net force on the balloon is

F=B-W=4159 N-2635 N=1524 N

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

W'=(m'+m)g=B

where m' is the additional mass. Re-arranging the equation for m', we find

m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

B=\rho_a V g

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

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