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Anna35 [415]
3 years ago
15

A balloon filled with helium gas has an average density of rhob = 0.27 kg/m3. The density of the air is about rhoa = 1.23 kg/m3.

The volume of the balloon is Vb = 0.084 m3. The balloon is floating upward with acceleration a.
Physics
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:34.84 m/s^2

   

Explanation:

Given

density of balloon \rho _{He}=0.27 kg/m^3

density of air \rho _a=1.23 kg/m^3

volume of balloon V=0.084 m^3

If balloon is rising then

F_b-mg=ma

where F_b=buoyant\ Force=\rho _a\times V\times g

mg=\rho _{He}\times V\times g=weight of gas

a=acceleration

\rho _a\times V\times g-\rho _{He}\times V\times g=\rho _{He}\times V\times a

\rho _a\times g-\rho _{He}\times g=\rho _{He}\times a

divide by \rho _{He}

\frac{\rho _a}{\rho _{He}}\times g-g=a

a=g(\frac{\rho _a}{\rho _{He}}-1)

a=g(\frac{1.23}{0.27}-1)

a=3.55\times 9.8

a=34.84 m/s^2

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A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
A crate is dragged up a ramp at constant speed. The work done on the system can be accounted for by:
Alex

Answer:

The work done on the system can be accounted for by;

Both E_g and E_{int}

Explanation:

The speed of the crate = Constant

Therefore, the acceleration of the crate = 0 m/s²

The net force applied to the crate, F_{NET} = 0

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The weight of the crate × The height the crate is raised = Gravitational Potential Energy = E_g

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<Continuation of the question>

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