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Crank
2 years ago
5

Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 970 km/hr above East Lansing. The

downward component of the earth's magnetic field at this place is
Physics
1 answer:
balandron [24]2 years ago
6 0

This question is incomplete, the complete question is;

Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 970 km/hr above East Lansing. The downward component of the earth's magnetic field at this place is 0.7 × 10⁻⁴ T. (DATA: Assume that the wingspan is 60 meters.)

Answer:

the average induced voltage between the tips of the wings of a Boeing 747 flying is 1.132 Volts

Explanation:

Given that;

Speed S = 970 KM/hr

downward component of the earth's magnetic field B = 0.7 × 10⁻⁴

wingspan 1 = 60min

Velocity V = (970 × 10³) / 3600 = 269.44 m/s

So Average Induced Voltage E = VBI

we substitute

E = 269.44 × (0.7 × 10⁻⁴) × 60

E = 1.132 Volts

Therefore the average induced voltage between the tips of the wings of a Boeing 747 flying is 1.132 Volts

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A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release
Scilla [17]

Answer:

(A) As it moves farther and farther from Q, its speed will keep increasing.

Explanation:

When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.

Mathematically:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}

where:

r = distance between the charges

\epsilon_0= permittivity of free space

By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as  the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.

7 0
3 years ago
What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a
Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

M=-\frac{f_o}{f_e}

Here, M is the magnification of an astronomical telescope, f_e is the focal length of the eyepiece lens and f_o is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put f_o=74 cm and f_e=2.4 cm in the above expression.

M=-\frac{74}{2.4}

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0
3 years ago
A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upo
Drupady [299]

Answer: 56.72 ft/s

Explanation:

Ok, initially we only have potential energy, that is equal to:

U =m*g*h

where g is the gravitational acceleration, m the mass and h the height.

h = 50ft and g = 32.17 ft/s^2

when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Then we have:

K = U

m*g*h = (m/2)*v^2

we solve it for v.

v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s

6 0
2 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
PLEASE HELP ME PLS PLS Imagine an object held at some height above the ground. It is released and falls toward the ground. Ignor
Stels [109]

Option e is true. The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

<h3>What is the law of conservation of energy?</h3>

According to the Law of conservation of energy. Although energy cannot be generated or destroyed, it may be transferred from one form to another.

The following statements are true;

a)All of its energy must be potential energy before it falls.

b)At the conclusion of its fall, all of its energy must be transformed to kinetic energy.

c)During its fall, the sum of its kinetic and potential energy must match the initial quantity of potential energy.

d)Total energy = Kinetic Energy + Potential Energy.

Hence, option e is correct.

To learn more about the law of conservation of energy, refer ;

brainly.com/question/2137260

#SPJ2

8 0
1 year ago
Read 2 more answers
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