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Nookie1986 [14]

1 year ago

11

Consider a blimp that can be approximated as a 3-m diameter, 8-m long ellipsoid and is connected to the ground. On a windless da

y, the rope tension due to the net buoyancy effect is measured to be 120 N. Determine the rope tension when there are 50 km/h winds blowing along the blimp (parallel to the blimp axis).

1 answer:

Marta_Voda [28]1 year ago

5 0

Step 1 of 6

Diameter of the blimp,

Length of the blimp,

Rope tension,

The velocity of the wind,

Step 2 of 6

The drag force acting on the blimp can be calculated using the formula given below,

(1)

Where, is the drag coefficient is the density of air.

A is the frontal area of the blimp.

V is the velocity of the wind.

Step 3 of 6

The frontal area of the parachute,

Here,

Substitute the value in the above equation.

The drag coefficient for the balloon is,

Step 4 of 6

From the table properties of air,

The density of air at 1atm pressure and is,

The velocity of air,

Step 5 of 6

Now, substitute all the known values in equation (1)

Step 6 of 6

Now, rope tension is the sum of rope tension due to the buoyancy effect and the force due to the wind blowing.

Learn more about the velocity at

brainly.com/question/4931057

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Acceleration of a refrigerator 30s after a person begins pushing it at a force of 400 N

ryzh [129]

Question: Find acceleration of a refrigerator 30s after a person begins pushing it at a force of 400 N, If the mass of the refrigerator is 10 kg.

Answer:

40 m/s²

Explanation:

Applying,

F = ma................Equation 1

Where F = Force applied to the refrigerator, m = mass of the refrigerator, a = acceleration of the refrigerator.

make a the subject of the equation

a = F/m............ Equation 2

From the question,

Given: F = 400 N, m = 10 kg

Substitute these values into equation 2

a = 400/10

a = 40 m/s²

6 0

2 years ago

1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol

torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0

3 years ago

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

$\Sigma F = F_{E}-W = 0$ (Eq. 1)

Where:

$F_{E}$ - Electrostatic force exerted on human, measured in Newton.

$W$ - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

$q\cdot E-m\cdot g = 0$

$q = \frac{m\cdot g}{E}$ (Eq. 2)

$E$ - Electric field, measured in Newtons per Coloumb.

$m$ - Mass, measured in kilograms.

$g$ - Gravity acceleration, measured in meters per square second.

$q$ - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that $m = 60\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $E = -150\,\frac{N}{C}$ , the charge that a 60-kg human must have to overcome weight is:

$q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }$

$q = -3.923\,C$

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

$F = \kappa \cdot \frac{q^{2}}{r^{2}}$ (Eq. 3)

Where:

$\kappa$ - Electrostatic constant, measured in Newton-square meter per square Coulomb.

$q$ - Electric charge, measured in Coulomb.

$r$ - Distance between two people, measured in meters.

If we know that $\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q = -3.923\,C$ and $r = 100\,m$ , then the force of repulsion between two people is:

$F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]$

$F = 13.851\times 10^{6}\,N$

The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

5 0

3 years ago

The electric potential at the dot in the figure is 3160 V. What is charge q?

PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

$V = \frac{kQ}{r}$

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

$V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V$

Lower left charge's potential:

$V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V$

Add the two, and subtract from the total EP at the point:

$3160 + 1123.75 - 2247.5 = 2036.25$

The remaining charge must have a potential of 2036.25 V, so:

$2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}$

5 0

2 years ago

What energy is calculated from an object’s mass, height, and the acceleration due to gravity? A. elastic kinetic B. elastic pote

Alex777 [14]

Answer: C. gravitational kinetic

Explanation: Gravitational potential energy is the energy calculated from an object's mass height and the acceleration due to gravity. The gravitational potential energy is the energy an object has as a result of the position of the object in a gravitational field.

3 0

3 years ago

Read 2 more answers