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Brrunno [24]
3 years ago
14

Why does the speed of sound vary depending on the medium through which it travels. in general?

Physics
1 answer:
avanturin [10]3 years ago
4 0
Because it is how fast it goes and it depends on the speed
You might be interested in
Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a
dusya [7]

Answer:

W=-21,870,000\ J

Explanation:

<u>Work and Kinetic Energy </u>

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

W=\Delta k=k_1-k_0

Since

\displaystyle k=\frac{mv^2}{2}

We have

\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}

W=0-21870000\ J

\boxed{W=-21,870,000\ J}

3 0
3 years ago
Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b
shepuryov [24]

Answer:

  • a. \Delta s ^2 = 8.0888 \ 10^{17} m^2
  • b. \Delta s ^2 = 3.0234 \ 10^{16} m^2
  • c. \Delta s ^2 = 3.0234 \ 10^{20} m^2

Explanation:

The spacetime interval \Delta s^2 is given by

\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2.

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2

\Delta \vec{x}^2 = 5,625 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2

\Delta (c t) ^ 2 = (899,377,374 \ m)^2

\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2

so

\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2

\Delta s ^2 = 8.0888 \ 10^{17} m^2

<h3>b.</h3>

\Delta \vec{x}^2 = (5 \ 10 \ m)^2

\Delta \vec{x}^2 = 2,500 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2

\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2

\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2

so

\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2

\Delta s ^2 = 3.0234 \ 10^{16} m^2

<h3>c.</h3>

\Delta \vec{x}^2 = (5 \ 10 \ m)^2

\Delta \vec{x}^2 = 2,500 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2

\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2

\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2

so

\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2

\Delta s ^2 = 3.0234 \ 10^{20} m^2

5 0
3 years ago
Inductors in parallel. Two inductors L1 = 1.31 H and L2 = 2.24 H are connected in parallel and separated by a large distance so
mote1985 [20]

Answer:

a) 0.83H

b) 3.22/N Henry

Explanation:

Given two inductors L1 = 1.31 H and L2 = 2.24 H connected in parallel, their equivalent inductance derivative is similar to that of resistance in parallel.

Derivation:

If the voltage across an inductor

VL = IXL

I is the current

XL is the inductive reactance

XL = 2πfL

VL = I(2πfL)

L is the inductance.

From the formula, I = V/2πfL

Given two inductors in parallel, different current will flow through them.

I1 = V/2πfL1 (current in L1)

I2 = V/2πfL2 (current in L2)

Total current I = I1+I2

I = V/2πfL1 + V/2πfL2

I = V/2πf{1/L1+1/L2}

V/2πfL = V/2πf{1/L1+1/L2}

1/L = 1/L1+1/L2 (equivalent inductance in parallel)

Given L1 = 1.31 H and L2 = 2.24

1/L = 1/1.31 + 1/2.24

1/L = 0.763 + 0.446

1/L = 1.209

L = 1/1.209

L = 0.83H

The equivalent inductance is 0.83H

b) Given similar inductors L = 3.22H in parallel, the equivalent inductance will be:

1/L = 1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

+1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

1/L = 10/3.22 (since that all have the same denominator)

L = 3.22/10

If N = 10, the generalization of 10 similar inductors in parallel will be:

L = 3.22/N Henry

4 0
3 years ago
Math Focus
zhenek [66]

Answer: 4.

Explanation:

Use formula v = d / t, where v = speed, d = distance and t = time.

v = 10 / 2.5

v = 4.

3 0
3 years ago
A resistor uses 100 w of power when connected to 120 v emf. What is the current through the same resistor when connected to a 22
cricket20 [7]

Answer:

Explanation:

Let the resistance of resistor be R .

Power of resistor  V² / R , where V is potential applied .

V² / R = 100

120² / R = 100

R = 120² / 100

= 144 ohm .

Now potential diff applied = 220 V

current = potential diff / resistance

= 220 / 144

= 1.53 A approx .

6 0
3 years ago
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