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3 years ago

A wave of infrared light has a speed of 6m/s and a wave length of 12 m. What is the frequency of this wave?

1 answer:

6 0

Frequency = speed / wavelength

(6 m/s) / (12 m) = 0.5 Hz.

That's not infrared light.
Infrared light waves move about 50 million times faster than that, and they're only about 0.00000007 as long as that.

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How do we usually hear things? Sound waves are made of ___ that pushes together ___.

denis-greek [22]

A then B, as everything is made of energy and energy pushes the electro magnetic waves into the brain.

7 0

3 years ago

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+138 determines the height of the rock abov

Likurg_2 [28]

Answer:

a) 138 ft

b) 4.62 s

c) 1.375 s

d) 168.25 ft

Explanation:

The height of a rock (thrown from the top of a bridge) above the level of water surface as it varies with time when thrown is given in the question as

h = f(t) = -16t² + 44t + 138

with t in seconds, and h in feet

a) The bridge's height above the water.

At t=0 s, the rock is at the level of the Bridge's height.

At t = 0,

h = 0 + 0 + 138 = 138 ft

b) How many seconds after being thrown does the rock hit the water?

The rock hits the water surface when h = 0 ft. Solving,

h = f(t) = -16t² + 44t + 138 = 0

-16t² + 44t + 138 = 0

Solving this quadratic equation,

t = 4.62 s or t = -1.87 s

Since time cannot be negative,

t = 4.62 s

c) How many seconds after being thrown does the rock reach its maximum height above the water?

At maximum height or at the maximum of any function, the derivative of that function with respect to the independent variable is equal to 0.

At maximum height,

(dh/dt) = f'(t) = (df/dt) = 0

h = f(t) = -16t² + 44t + 138

(dh/dt) = (df/dt) = -32t + 44 = 0

32t = 44

t = (44/32)

t = 1.375 s

d) What is the rock's maximum height above the water?

The maximum height occurs at t = 1.375 s,

Substituting this for t in the height equation,

h = f(t) = -16t² + 44t + 138

At t = 1.375 s, h = maximum height = H

H = f(1.375) = -16(1.375²) + 44(1.375) + 138

H = 168.25 ft

Hope this Helps!!!

3 0

3 years ago

1. A student pushes horizontally on a book with a force of 2 N, which causes the book to slide at a constant velocity. What is t

Cloud [144]

Answer:

The horizontal forces are equal in size and opposite in direction. They are balanced, so the horizontal resultant force is zero. This means that there is no horizontal acceleration just a horizontal constant speed. If a submarine is not moving horizontally, then there are no horizontal forces.

I hope it helps.

8 0

2 years ago

If you drop a ball off a cliff, it starts out at 0 m/s. after 1 s, it will be traveling at about 10 m/s. if air resistance is re

anastassius [24]

The ball dropped of a cliff will have a final velocity of 19.6 m/s at 2s

The free fall formula and the procedure we will use is:

fv = g* t

Where:

fv = final velocity
g = acceleration due to gravity
t = time taken
h = height traveled

Information of the problem:

g = 9.8 m/s².
t = 2 s
vf = ?

Applying final velocity formula we get:

fv = g* t

fv = 9.8 m/s² * 2 s

fv = 19.6 m/s

It is when the object or mobile falls from a height (h) with a positive acceleration equal to the gravity, describing a vertical rectilinear travel.

Learn more about free fall at brainly.com/question/11257529

#SPJ4

5 0

2 years ago

The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

Evgen [1.6K]

Question Continuation

Derive an expression for x in terms of m, M, and D. b) If the net force is zero a distance ⅔D from the planet, what is the ratio R of the mass of the planet to the mass of the moon, M/m?

Answer:

a. x = (D√M/m)/(√M/m + 1)

b. The ratio R of the mass of the planet to the mass of the moon=4:1

Explanation:

Given

m = Mass of moon

M = Mass of the planet

D = Distance between the centre of the planet and the moon

Net force = 0

Let Y be a point at distance x from the planet

Let mo = mass at point Y

Derive an expression for x in terms of m, M and D.

Formula for Gravitational Force is

F = Gm1m2/r²

Y = D - x

Force on rest mass due to mass M (FM) =Force applied on rest mass due to m (Fm)

FM = G * mo * M/x²

Fm = G * mo * m/Y²

Fm = G * mo * m/(D - x)²

FM = Fm = 0 ------ from the question

So,

G * mo * M/x² = G * mo * m/(D - x)² ----- divide both sides by G * mo

M/x² = m/(D - x)² --- Cross Multiply

M * (D - x)² = m * x²

M/m = x²/(D - x)² ---_ Find square roots of both sides

√(M/m) = x/(D - x) ----- Multiply both sides by (D - x)

(D - x)√(M/m) = x

D√(M/m) - x√(M/m) = x

D√(M/m) = x√(M/m) + x

D√(M/m) = x(√(M/m) + 1) ------- Divide both sides by √M/m + 1

x = (D√M/m)/(√M/m + 1)

b. Here x = ⅔D

FM = G * mo * M/x²

Fm = G * mo * m/(D - x)²

FM = Fm

G * mo * M/x² = G * mo * m/(D - x)² ----- divide both sides by G * mo

M/x² = m/(D - x)² --- (Substitute ⅔D for x)

M/(⅔D)² = m/(D - ⅔D)²

M/(4D/9) = m/(⅓D)²

9M/4D = m/(D/9)

9M/4D = 9m/D ---- Divide both side by 9/D

M/4 = m

M = 4m

M/m = 4

M:m = 4:1

So, the ratio R of the mass of the planet to the mass of the moon=4:1

3 0

3 years ago