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Alla [95]
4 years ago
10

PLEASE HELP ASAP: Wave Y and wave Z are two types of electromagnetic waves traveling through a vacuum. Suppose the frequency of

wave Y is about three times greater than the frequency of wave Z. Which best compares the energy transferred by the two waves? It is the same for each wave because the amount of energy transferred is constant in all types of electromagnetic waves. Wave Y transfers about one-third the amount of energy as wave Z because energy and frequency have an inverse relationship. Wave Y transfers about three times the amount of energy as wave Z because energy and frequency have a direct relationship. It is zero for each wave because when waves travel through a vacuum, they do not transfer energy.
Physics
2 answers:
skad [1K]4 years ago
9 0

Answer:

c

Explanation:

c

slega [8]4 years ago
4 0

Frequency and energy have a direct relationshipWave Y transfers about three times the amount of energy as wave Z because energy and frequency have a direct relationship. Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and
ira [324]

Complete question is;

a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?

b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be

Answer:

A) V_t = 18 m/s

B) V_t = 10.39 m/s

Explanation:

Formula for terminal speed is given by;

V_t = √(2mg/(DρA))

Where;

m is mass

g is acceleration due to gravity

D is drag coefficient

ρ is density

A is Area of object

A) Now, for sphere 1,we have;

m = 1 kg

V_t = 6 m/s

g = 9.81 m/s²

Now, making D the subject, we have;

D = 2mg/((V_t)²ρA))

D = (2 × 1 × 9.81)/(6² × ρA)

D = 0.545/(ρA)

For sphere 2, we have mass = 9 kg

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]

V_t = 18 m/s

B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.

Thus;

Area of sphere 2 = 3A

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]

V_t = 10.39 m/s

5 0
4 years ago
A particular spring has a spring constant of 25 N/m.
kotegsom [21]

Answer:

<h2>2.8×10^-3</h2>

e.p.e =  \frac{1}{2} k {x}^{2}   \\   = \frac{1}{2}  \times 25 \times (0.015m)^{2}  \\  = 0.0028125

<u>0.0028125N</u><u> </u>

8 0
3 years ago
A 0.10 newton spring toy with a spring constant of 160 newtons per meter is compressed 0.05 meter before it is launched. When re
forsale [732]

Answer:

(1) V = 0.2 J (2) 0.05J

Explanation:

Solution

Given that:

K = 160 N/m

x = 0.05 m

Now,

(1) we solve for the  initial potential energy stored

Thus,

V = 1/2 kx² = 0.5 * 160 * (0.05)²

Therefore V = 0.2 J

(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.

By using the energy conversion, we have the following

ΔV = mgh

=(0.1/9.8) * 9.8 * 1.5 = 0.15J

The internal energy = 0.2 -0.15

=0.05J

8 0
3 years ago
A 2.0 g identification reflector glued to one end of a helicopter rotor is spinning at a tangential velocity of 2093 m/s. The re
katrin2010 [14]
372863 juusnjus bhhhanbubhgajhus
7 0
3 years ago
A 2150 kg satellite used in a cellular telephone network is in a circular orbit at a height of 780 km above the surface of the e
Tom [10]

Answer:

a)F=16741.9N

b)\frac{F}{W}=0.795

Explanation:

The gravitational force on the satellite is calculated with Newton's Gravitation Law:

F=\frac{GMm}{r^2}

Where M=5.97\times10^{24}kg is Earth's mass, m=2150kg is the satellite mass, r=R+h is the distance between their centers, where h=780000m is the height of the satellite (from Earth's surface) and R=6371000m is Earth's radius, and G=6.67\times10^{-11}Nm^2/kg^2 is the gravitational constant.

a) With these values we then have:

F=\frac{GMm}{r^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)(2150kg)}{(6371000m+780000m)^2}=16741.9N

b) And the fraction this force is of the satellite’s weight <em>W=mg</em> is:

\frac{F}{W}=\frac{GMm}{mgr^2}=\frac{GM}{gr^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)}{(9.8m/s^2)(6371000m+780000m)^2}=0.795

5 0
4 years ago
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