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icang [17]
3 years ago
10

In the lab activity, you will examine sound waves as they are emitted from a moving source. Predict what will happen to the soun

d waves when the sound source is in motion. Record your prediction (hypothesis) as an “if then” statement. (For example: If you select the GO button, then the train will move)
Physics
2 answers:
irina1246 [14]3 years ago
6 0
<span>As a sound source gets closer, both the volume and the pitch of the sound increased. Then, as the sound source passed by you, both the volume and the pitch of the sound decreased.
 Hope this helps</span>
Serjik [45]3 years ago
5 0

Answer:

Explanation:

If the source is the same, the characteristic sound waves when they are emitted do not change (wave speed), but if the source is in relative motion with the observer, a change in the wavelength must be seen by this relative movement, such as the speed is the same this will be reflected in a change in the frequency heard by the observer

This effect is called the Doppler effect for a moving source is described by the expression, and

     f ’= [(v / (v± vs)] f

With v the speed of sound, vs the speed of the source, f the frequency emitted and f ’the frequency heard, where the sign corresponds to the source away (-) or the source approaching (-)

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A 1400 kg car is moving at 12 m/s when the driver stops a car what increase in the thermal energy of the car and it surroundings
butalik [34]
KE = 1/2 mv^2

in this case, the initial kinetic energy which is converted to heat is 

KE = 1/2 1400 (12)^2

KE = 100,800 J
5 0
3 years ago
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Planet x has a mass of 4x1022 kg and a radius of 6x105 m What lise the grav ritational field strength in the surface of planet X
SCORPION-xisa [38]

Answer:

  g ≈ 7.4 m/s²

Explanation:

The acceleration due to gravity on planet XX is ...

  g = GM/r² = (6.67·10^-11 × 4·10^22)/(6·10^5)^2

  g ≈ 7.4 m/s²

6 0
3 years ago
Question 1 of 10
Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field E at a distance R from a charge Q is given by

E = k\dfrac{Q}{R^2}

where k = 9*10^9Nm/C is the coulomb's constant.

Now, in our case

R = 0.0075m

Q = 0.0035C;

therefore,

E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}

\boxed{E = 5.6*10^{11}N/C.}

which is choice C from the options given<em> (at least it resembles it).</em>

6 0
3 years ago
A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

#SPJ1

6 0
2 years ago
A 2kg object is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete rev
Travka [436]

Answer:

a) 37.70 m/s

b)710.6 m/s²

Explanation:

Given that ;

Mass of object = 2 kg

Radius of the motion = 2m

Frequency of motion = 3 rev/s

The formula to apply is;

v= 2πrf   where v is linear speed

v = 2×π×2×3 =12π = 37.70 m/s

Centripetal acceleration is given as;

a= 4×π²×r×f²  

a= 4×π²×2×3²

a=710.6 m/s²

5 0
3 years ago
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