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scZoUnD [109]
3 years ago
5

During constructive interference two waves travel through the same medium and the effects of the first wave are ___ the effects

of the second wave
Physics
1 answer:
Ymorist [56]3 years ago
7 0

Answer:  

During constructive interference occurs at any location along the medium  where two inferring waves are in the same direction. In this case both waves have upward displacement. Consequently the medium has an upward displacement has that is greater than the displacement of inferring pulses.


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5lbs is greater.
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3 years ago
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The current theory of the structure of the
Mariana [72]

Answers:

a) 2.82(10)^{21} kg

b) 1410 J

c) 36.62 m/s

Explanation:

<h3>a) Mass of the continent</h3>

Density \rho  is defined as a relation between mass m and volume V:

\rho=\frac{m}{V} (1)

Where:

\rho=2720 kg/m^{3} is the average density of the continent

m is the mass of the continent

V is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}

Finding the mass:

m=\rho V (2)

m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3}) (3)

m=2.82(10)^{21} kg (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy K is given by the following equation:

K=\frac{1}{2}mv^{2} (5)

Where:

m=2.82(10)^{21} kg is the mass of the continent

v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s is the velocity of the continent

K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2} (6)

K=1410 J (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass m=77 kg and the same kinetic energy as that of the continent 1413 J, we can find its velocity by isolating v from (5):

v=\sqrt{\frac{2 K}{m}} (6)

v=\sqrt{\frac{2 (1413 J)}{77 kg}}

Finally:

v=36.62 m/s This is the speed of the jogger

5 0
3 years ago
As you learned in Part B, a non-burning helium core surrounded by a shell of hydrogen-burning gas characterizes the subgiant sta
Gennadij [26K]

Answer:

E- The star becomes a red giant (LATEST STAGE)

F- The surface of the star becomes brighter and cooler

C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand

A- The shell of hydrogen surrounding the star's nonburning helium core ignites.

D- The star's non burning helium core starts to contract and heat up

B- Pressure in the star's core decreases (EARLIEST STAGE)

(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.

6 0
3 years ago
A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun
cluponka [151]

Answer:Highest frequency  =618.89Hz

Lowest frequency=582.22Hz

Explanation:

 The linear velocity of a sound generator  is related to angular velocity and is given as

Vs = rω  where

r = the radius of circular path = 1.0 m

ω is the angular velocity of the sound generator. = 100 rpm

1 rev/min = 0.10472 rad/s

100rpm =10.472 rad/ s

Vs = rω

= 1m x 10.472rad/ s=  10.472m/s

A) Highest frequency  heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,

f max = (v/ v-vs) fs

Where , v is the speed of the sound in air at 20 degrees celcius =

343 metres per second

vs is the linear velocity of the sound generator=10.472m/s

fs is the frequency of the sound generator= 600 Hz

f max = (343/ 343 - 10.472) x 600

=343/332.528) x600

=618.89Hz

B) Lowest frequency  heard by a student in the classroom = Minimum frequency

f min = (v/ v+vs) fs

(343/ 343 + 10.472) x 600

=343/353.472) x 600

=582.22hz

5 0
3 years ago
If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?
steposvetlana [31]

Hi there!

\large\boxed{\text{B) 20 meters}}

We know that:

E_T = U + K

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².

50J = 10(10)m

m = 1/2 kg

Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

3 0
2 years ago
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