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Aleksandr-060686 [28]
3 years ago
15

It is determined that a certain light wave has a wavelength of 3.012 x 10-12 m. The light travels at 2.99 x 108 m/s. What is the

frequency of the light wave? (Round your answer to three significant figures.)
*Answer in HZ units
Physics
1 answer:
gayaneshka [121]3 years ago
3 0

Answer:

id.k

Explanation:

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An object completes one and half revolution of a circle of radius R calculate the displacement and distance
Papessa [141]

Answer:

Distance is path length covered by particle. When particle moves along half circle, it covers half the circumference therefore distance covered is (2×pi×r)/2 = pi× r. ... Hence displacement is equal to diameter or 2 times the radius of circle.

8 0
3 years ago
How fast would the International Space Station (ISS) have to travel to maintain a circular orbit a distance of 1400 km above the
professor190 [17]

Answer:

The International Space Station move at 7.22 km/s.

Explanation:

Orbital speed of satellite is given by  v=\sqrt{\frac{GM}{r}}, where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.

r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m

G = 6.673 x 10⁻¹¹ Nm²/kg²

M = 5.98 x 10²⁴ kg

Substituting

              v=\sqrt{\frac{6.673\times 10^{-11}\times 5.98\times 10^{24}}{7.75\times 10^6}}=7223.86m/s=7.22km/s

  The International Space Station move at 7.22 km/s.      

4 0
3 years ago
A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 3.30 seconds af
disa [49]

Answer:

Explanation:

Given

Cannon is fired with a velocity of u=72.50\ m/s

Using Equation of motion

y=ut+\frac{1}{2}at^2

where

y=displacement

u=initial\ velocity

a=acceleration

t=time

after time t=3.3 s

y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2

y=239.25-53.36

y=185.89\ m

So after 3.3 s cannon ball is at a height of 185.89 m

6 0
3 years ago
A QL = -26 μC charge is placed on the x-axis at x = - 0.2 m. A QR = 26 μC charge is placed at x = +0.2 m. (for all answers below
Kaylis [27]
<span>You are given a QL = -26 μC charge that is placed on the x-axis at x = - 0.2 m and a QR = 26 μC charge that is placed at x = +0.2 m. The answers are:

The x-component of the electric field at x = 0 m and y = 0.2 m is 3.
The y-component of the electric field at x = 0 m and y = 0.2 m is 2.

</span>
4 0
3 years ago
6. A 145-g baseball moving 30.5 m/s strikes a stationary 5.75-kg brick resting on small rollers so it moves without significant
Sindrei [870]

Answer:

Explanation:

a )

momentum of baseball before collision

mass x velocity

= .145 x 30.5

= 4.4225 kg m /s

momentum of brick after collision

= 5.75 x 1.1

= 6.325 kg m/s

Applying conservation of momentum

4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.

v = - 13.12 m / s

b )

kinetic energy of baseball  before collision = 1/2 mv²

= .5 x .145 x 30.5²

= 67.44 J

Total kinetic energy before collision = 67.44 J

c )

kinetic energy of baseball after collision = 1/2 x .145 x 13.12²

= 12.48 J .

 kinetic energy of brick after collision

= .5 x 5.75 x 1.1²

= 3.48 J

Total kinetic energy after collision

= 15.96 J

3 0
3 years ago
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