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sineoko [7]
3 years ago
13

  If a string vibrates at the fundamental frequency of 528 Hz and also produces an overtone with a frequency of 1,056 Hz, this o

vertone is the 
*A. *second harmonic.
*B. *first harmonic.
*C. *fourth harmonic.
*D. *third harmonic.

14.   The _______ of a sound wave is defined as the amount of energy passing through a unit area of the wave front in a unit of time. 
*A. *amplitude
*B. *compression
*C. *intensity
*D. *frequency
Physics
1 answer:
Leto [7]3 years ago
6 0

your answer is *A. *second harmonic.

i just took the test

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I NEED HELP!!! PLS HELP ME MARK U BRAINLIEST!!
FinnZ [79.3K]

Answer:

The answer is X

Explanation:

Cause the highest points will most likely have the most potential energy

8 0
3 years ago
Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros
kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

v^2-u^2=2gh

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g=-9.8 m/s^2 is the acceleration of gravity

Solving for u,

u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s

The time needed to reach the maximum height can now be found by using the equation

v=u+gt

Solving for t,

t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

v'^2-u^2=2gh'

we find

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s

So the time to go from h' to h is

\Delta t = t-t'=0.52-0.32=0.20 s

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

\Delta t=2\cdot 0.20 = 0.40 s

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

v'^2-u^2=2gh'

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

\Delta t = 2\cdot 0.04 =0.08 s

8 0
4 years ago
A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

4 0
2 years ago
PLEASE HELP! Can you do the write about it! Will thank and try to mark brainiest!
goblinko [34]
Move the objects faster to get more friction.
8 0
3 years ago
Read 2 more answers
Light with a wavelength of 400 nm strikes the surface of cesium in a photocell, and the maximum kinetic energy of the electrons
Firdavs [7]

Answer:

The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

Explanation:

Given that,

Wavelength = 400 nm

Energy E=1.54\times10^{-19}\ J

We need to calculate the longest wavelength of light that is capable of ejecting electrons from that metal

Using formula of energy

E = \dfrac{hc}{\lambda}

\lambda=\dfrac{hc}{E}

Put the value into the formula

\lambda=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.54\times10^{-19}}

\lambda=1292\times10^{-9}\ m

\lambda=1292\ nm

Hence, The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

8 0
3 years ago
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