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aivan3 [116]
3 years ago
10

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

is filled with helium gas until its volume is 323 m3.
Assume the density of air is 1.20 kg/m3 and the density of helium is 0.179 kg/m3.

(a) Calculate the buoyant force (in N) acting on the balloon. (Enter the magnitude. Round your answer to at least three significant figures.)

(b) Find the net force (in N) on the balloon and determine whether the balloon will rise or fall after it is released. magnitude N direction rise D

(c) What additional mass (in kg) can the balloon support in equilibrium? (If the balloon will fall after it is released, enter 0.)
Physics
1 answer:
Orlov [11]3 years ago
4 0

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

F= ma (\rightarrow Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

\rho = \frac{m}{V} \Rightarrow m = \rho V

Part A) The buoyant force acting on the balloon is given as

F_b = ma

As mass is equal to the density and Volume and acceleration equal to Gravity constant

F_b = \rho V g

F_b = 1.2*323*9.8

F_b = 3798.5

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

F_{net} = F_b -W

F_{net} = F_b -(mg+\rho_H Vg)

F_{net} = 3798.5-(9.8*225*9.8*0.179*323)

F_{net} = 1030N

PART C) The additional mass that can the balloon support in equilibrium is given as

F_{net} = m' g

m' =\frac{F_{net}}{g}

m' = \frac{1030}{9.8}

m' = 105Kg

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charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
2 years ago
A person drives for 1.00 h at 10.0 km/h and 2.00 h at 20.0 km/h. What is their average speed, in SI units?
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I believe it is b if i’m correct
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The wavelength of green light is about the size of an atom. (T/F)
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Explanation:

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A 25 kg ball is thrown into the air when thrown it's going 10 m/s calculate how high it travels
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On Earth, the acceleration of gravity is 9.8 m/s² downward.
So any object with only gravity acting on it gains 9.8 m/s of
downward speed every second.

If the rock starts out moving upward at 10 m/s, then it will
continue upward for only  (10/9.8) = 1.02 second, before
it stops rising and starts falling.

Its average speed during that time is  (1/2) (10 + 0) = 5 m/s .

At an average speed of 5 m/s for 1.02 sec,
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A man at point A directs his rowboat due north toward point B, straight across a river of width 100 m. The river current is due
prohojiy [21]

Answer:

1.35208 m/s

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Speed of the boat = 0.75 m/s

Distance between the shores = 100 m

Time = Distance / Speed

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Time taken by the boat to get across is 133.33 seconds

Point C is 150 m from B

Speed = Distance / Time

Speed=\frac{150}{\frac{100}{0.75}}=1.125\ m/s

Velocity of the water is 1.125 m/s

From Pythagoras theorem

c=\sqrt{0.75^2+1.125^2}\\\Rightarrow c=1.35208\ m/s

So, the man's velocity relative to the shore is 1.35208 m/s

3 0
3 years ago
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