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3 years ago

15

A plasma wave moving through a plasma has a frequency of 109 Hz and a speed of 3.0 × 107 m/s. What is the wavelength of this wav

e?

1 answer:

tigry1 [53]3 years ago

8 0

Answer:0.27*10^{6} (m)

Explanation:

with a wave we know that speed= wavelenght* frecuency so

wavelenght= 3*10^{7} / 109 = 0.27*10^{6} (m)

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A teacher explains that a scientist named Boyle did many experiments to determine what the relationship is between the volume an

lina2011 [118]

<span>Boyle's Law, because it describes what will happen in the relationship between the pressure and volume of the gas.</span>

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3 years ago

Two Charged spheres of 6.4 nC charge are 16 cm apart. What is the magnitude of the field in middle? ( n = 10^-9, 100 cm = 1 m)

emmainna [20.7K]

Answer:

zero

Explanation:

q = 6.4 nC = 6.4 x 1 0^-9 C

d = 16 cm = 0.16 m

r = 16 / 2 = 8 cm = 0.08 m

Electric field at P due to the charge placed at A

Ea = k q / r^2

Ea = ( 9 x 10^9 x 6.4 x 10^-9) / (0.08 x 0.08) Towards right

Ea = 9000 Towards right

Electric field at P due to the charge placed at B

Eb = k q / r^2

Eb = ( 9 x 10^9 x 6.4 x 10^-9) / (0.08 x 0.08) Towards left

Eb = 9000 Towards left

The magnitude of electric field is same but teh direction is opposite, so the resultant electric field at P is zero.

3 0

3 years ago

What type of energy is released when a firecracker explodes

Ludmilka [50]

-- Sound energy. That's how you hear it explode.

-- Light energy. That's how you see it in the sky.

-- Heat energy. That's why you get burned if sparks fall on you.

-- Potential energy, as all the pieces fall to the ground.

These were all stored in it as chemical energy before it exploded.

5 0

4 years ago

Read 2 more answers

1. *A car is going over the top of a hill whose curvature approximates a circle of radius 200 m. At

Greeley [361]

Answer:

The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight is approximately 19.81 m/s

Explanation:

The given parameters are;

The curvature of the hill, r = 200 m

Due to the velocity, v, the occupants weight = 20% less than the normal weight

The outward force of an object due to centripetal (motion) force is given by the following equation;

$F_c = \dfrac{m \times v^2}{r}$

Where;

r = The radius of curvature of the hill = 200 m

Given that the weight of the occupants, W = m × g, we have;

$F_c = 0.2 \times W = 0.2 \times m \times g$

$\therefore 0.2 \times m \times g = \dfrac{m \times v^2}{r}$

v = √(0.2 × g × r)

By substitution, we have;

v = √(0.2 × 9.81 × 200) ≈ 19.81

The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight ≈ 19.81 m/s.

3 0

3 years ago

If a mass of 76 kg acts downward 0.38 m from the axis of rotation on one end of a board and another force of 129 N also acts dow

Semenov [28]

Answer:

y = 2.196 m

Explanation:

Mass, m = 76 kg

distance from axis of rotation, x = 0.38 m

Second Force, F = 129 N

moment arm of the second force, y = ?

Now, equating moments for the equilibrium

So,

m g × x = F x y

76 x 0.38 x 9.81 = 129 x y

$y = \dfrac{76\times 0.38\times 9.81}{129}$

y = 2.196 m

Hence, the length of the moment arm is equal to 2.196 m.

6 0

3 years ago