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3 years ago

What is the approximate weight of a 20-kg cannonball on earth

Physics

2 answers:

djverab [1.8K]3 years ago

7 0

Answer:

196 N

Explanation:

The weight of an object is given by

$W=mg$

where m is the mass of the object, and g is the gravitational acceleration on the planet.

In this example, we have an object with mass m=20 kg, and the value of the acceleration due to gravity on Earth is g=9.81 m/s^2, therefore if we use the above equation we find the weight of the cannonball:

$W=mg=(20 kg)(9.8 m/s^2)=196 N$

laiz [17]3 years ago

6 0

20 times 9.8 = 196 N

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Consider a monochromatic electromagnetic plane wave propagating in the x direction. At a particular point in space, the magnitud

Scrat [10]

Answer:

a) S = 2.35 10³ J/m²2 ,

b)and the tape recorder must be in the positive Z-axis direction.

the answer is 5

c) the direction of the positive x axis

Explanation:

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S = 1 /μ₀ E x B

if we use that the fields are in phase

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we substitute

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let's calculate

s = 941 2 / (4π 10⁻⁷ 3 10⁸)

S = 2.35 10³ J/m²2

b) the two fields are perpendicular to each other and in the direction of propagation of the radiation

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C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis

5 0

3 years ago

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

3 years ago

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Answer:

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Answer:

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Explanation:

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