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taurus [48]
3 years ago
14

What instrument can be used to show the shape of a sound wave?

Physics
1 answer:
mafiozo [28]3 years ago
7 0
"Oscilloscope"  can be used to show the shape of a sound wave

Hope this helps!
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What happens when a light bulb is disconnected in a parralel circuit? ​
VARVARA [1.3K]

Answer:

Hope i could help!

Explanation:

so all but one light could be burned out, and the last one will still function.

5 0
3 years ago
Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and
dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)


8 0
3 years ago
Read 2 more answers
Gravity is dependent on which of the two factors? mass and weight distance and weight mass and force mass and distance
andre [41]
Gravity = m₁m₂G / r²

G constant
m mass
r distance
8 0
3 years ago
Read 2 more answers
How many Mars-sized planets can fit in Jupiter
Maru [420]

Answer:

You could put over six planets the size of Mars inside the Earth. The largest planet in our Solar System, Jupiter's size is astounding. Jupiter has a volume of 1.43 x 1015 cubic kilometers. To show what this number means, you could fit 1321 Earths inside of Jupiter

Explanation:

7 0
3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
Inessa05 [86]

Answer:

a) 4.40 s

b) 2.20 s

Explanation:

Given parameters are:

At constant power  ,

initial speed of the car, v_0=0

final speed of the car, v=32 mph

At full power,

initial speed of the car, v_0=0

final speed of the car, v=64 mph

a)

At constant power, KE = \frac{1}{2} mv^2

At full power, KE = \frac{1}{2} m(2v)^2

So KE_f = 4KE_i

So, time to reach 64 mph speed is 4 times more than the initial time

t = 4*1.10 =4.40 s

b)

v=v_0+at\\a=\frac{v-v_0}{t}=\frac{32-0}{1.1/3600}=104727.27 miles/hours^2

For final 64 mph speed,

v=v_0+at\\t=\frac{v-v_0}{a}=\frac{64-0}{104727.27} = 6.111*10^{-4} hours = 6.111*10^{-4}*3600=2.20 s

7 0
3 years ago
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