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2 years ago

9

A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi

tional weight of the balloon material itself. The balloon material has a mass of 60 g/m2. Ambient air is at 25 °C and 1 atm. The hot air inside the balloon is at 70 °C and 1 atm. Assuming that the balloon has a perfectly spherical shape, what balloon diameter will just support the total weight? Neglect the size of the hot air inlet vent and assume an ideal gas law (P = rhoRT) for the air.

1 answer:

RSB [31]2 years ago

3 0

Answer:

r = 4.44 m

Explanation:

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

B = ρ g V

Now let's use Newton's equilibrium relationship

B - W = 0

B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

σ = W / A

W = σ A

The area of a sphere is

A = 4π r²

W = W₁ + σ 4π r²

The volume of a sphere is

V = 4/3 π r³

Let's replace

ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

P V = n RT

P = ρ RT

ρ = P / RT

P / RT g 4/3 π r³ - σ 4 π r² = W₁

r² 4π (P/3RT r - σ) = W₁

Let's replace the values

r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

r² (11.81 r -0.060) = 13000 / 4pi

r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

r = 4.44 m

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Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and

scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

7 0

2 years ago

Dentists' chairs are examples of hydraulic-lift systems. If a chair weighs 1400 N and rests on a piston with a cross-sectional a

NeX [460]

Answer:

Force applied to smaller cross section is

= 82.63 N

Explanation:

As we know

$F_2 A_1 = F_1 A_2$

where $F1, F2$ signifies the weight of the two chair in a hydraulic-lift system

And $A_1, A_2$ signifies the area of the two respective chairs in a hydraulic-lift system

Given -

$F2=1400$ N

$A1 =1220$ Square centimeter

$A_2 = 72$ Square centimeter

Substituting the given values in above equation, we get -

$1400 * 72 = F1 * 1220\\F2 = 82.63$

Force applied to smaller cross section is

= 82.63 N

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2 years ago

While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$

$m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sin$ Ф

60 ·8 + 0 = (60 + 80) $V_{f}sin$ Ф

480 = 140 $V_{f} sin$ Ф................. (I)

y-axis component form (+y north);

$P_{Mix}$ + $p_{Pix}$ + $j_{x}$ = $P_{Mfx}$ + $P_{pfx}$

$m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos$ Ф

0 + 80.9 = (60 + 80) $V_{f}cos$ Ф

720= 140 $V_{f}cos$ Ф

140Vf= $\frac{720}{cos}$ Ф......................................(2)

Substituting (2) into (1) to give the angle;

480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

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2 years ago

To understand the forces between a bar magnet and 1. a stationary charge, 2. a moving charge, and 3. a ferromagnetic object. A b

Brums [2.3K]

Answer:

Explanation:

1. If the charge is stationary and it is not moving, then the magnetic force will be zero, since the velocity is 0m/s

From the formula of magnetic force

F = q(V×B)

And since V is 0, then, the magnetic force is zero

F = 0N,

No torque at all.

2. The moving charge interacts with the fixed magnet. The force between them is at a maximum when the velocity of the charge is perpendicular to the magnetic field.

From the formula of magnetic force

F = q( V×B)

V × B = |V|•|B| Sinθ

So, Sinθ has its highest value at 90° in the range of 0-360°

So, the force will have a maximum value when the moving charge is perpendicular to the magnetic field

Now, since the magnetic field is in the positive y-direction, for the force to be maximum the electron must move the positive x-direction.

So the force will be in the positive z-direction(since i×j= k).

3. Now, the charge is replaced by an electrically neutral piece of initially unmagnetized soft iron (e.g. nail) that is not moving. This result into the magnetic experience a force due to the iron fillings attracting its north pole.

3 0

2 years ago

A person stands a distance R from a door's hinges and pushes with a force F directed perpendicular to its surface. By what facto

Zina [86]

Answer:

a)

4 times

b)

2 times

c)

0.5 times

d)

0.25 times

Explanation:

$F_{a}$ = Applied force by the person = F

$r$ = distance from the hinge = R

Torque is given as

$\tau = rF_{a}\\\tau = R F$

(a)

$F_{a}$ = Applied force by the person = 2F

$r$ = distance from the hinge = 2R

New Torque is given as

$\tau_{new} = rF_{a}\\\tau_{new} = (2R) (2F)\\\tau_{new} = 4RF\\\tau_{new} = 4 \tau$

So Torque becomes 4 times

b)

$F_{a}$ = Applied force by the person = F

$r$ = distance from the hinge = 2R

New Torque is given as

$\tau_{new} = rF_{a}\\\tau_{new} = (2R) (F)\\\tau_{new} = 2RF\\\tau_{new} = 2 \tau$

So Torque becomes 2 times

c)

$F_{a}$ = Applied force by the person = (0.5)F

$r$ = distance from the hinge = R

New Torque is given as

$\tau_{new} = rF_{a}\\\tau_{new} = (R) (0.5)(F)\\\tau_{new} = (0.5)RF\\\tau_{new} = (0.5) \tau$

So Torque becomes 0.5 times

d)

$F_{a}$ = Applied force by the person = (0.5)F

$r$ = distance from the hinge = (0.5)R

New Torque is given as

$\tau_{new} = rF_{a}\\\tau_{new} = (0.5)(R) (0.5)(F)\\\tau_{new} = (0.25)RF\\\tau_{new} = (0.25) \tau$

So Torque becomes 0.25 times

6 0

2 years ago