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Furkat [3]
3 years ago
7

"Calculate the speed of this sound wave by recording how much time it takes to travel 5 meters. Use the (old school) velocity eq

uation: v = d/t Show units in your work. Answer: 292.39 m/s= 5 meters/.0171 secs." and here's the question I need help with: "Based on the speed you measured above, how long would it take for you to hear thunder if you observed lightning and it was seen to be 1620 meters away?"
Physics
1 answer:
lianna [129]3 years ago
7 0
1) sound velocity reported by you : 292.39 m /s

2) time to travel 1620m at that velocity: t = d / v = 1620 m / 292.39 m/s = 5.54 s, since the moment the sound wave started.

3) You might wanted to tell the time since you watched the lightning.

Then you can calculate the time since the lighting was generated,1620 m away from you, until you saw it, using the speed of light:

 speed of light = 3*10^8 m/s => t = 1620 m / (3*10^8m/s) =0.0000054 s

Then, this time is completely neglectible, and yet the answer is 5.54 s, as calculated in the step 2.
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A period of the periodic table ends when the highest energy level of an elements atoms has __ electrons
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This is dependent on how many shells/layers/energy levels the element has. The first shell can only hold 2 electrons however every shell beyond that can hold 8 electrons
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An ideal solenoid having a coil density of 5000 turns per meter is 10 cm long and carries a current of 4.0
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The rule that is used to get the strength of magnetic field at the center of solenoid (B) is:
B = <span>µ x n x I where:
</span>µ is the permeability of the medium where the solenoid is based. In this problem, the medium is air which means that µ = <span>µ </span><span>o = 4 pi x 10^-7 Tm/A
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n is the number of turns per unit length (5000 turns)

Substituting in the mentioned equation, we find that:
B = 4 x 3.14 x 10^-7 x 5000 x 4 = 25.132 mT
5 0
3 years ago
Six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then
Nataly_w [17]

Answer:

The option (b) is correct.

Explanation:

The expression for the power in terms of work done is as follows;

P=\frac{W}{t}

Here, W is the work done, t is the time taken and P is the power.

According to the given problem, six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then determined.

The expression for the equivalent resistance in the series combination is as follows;

R_{eq}=R_{1}+R_{2}......+R_{6}

Put R_{1},R_{2}......,R_{6}=R.

R_{eq}=R+R+R+R+R+R

R_{eq}=6R

It is given in the problem that a seventh resistor is added (in series).

R_{eq}=R_{1}+R_{2}......+R_{7}

Put R_{1},R_{2}......,R_{7}=R.

R_{eq}=R+R+R+R+R+R+R

R_{eq}=7R

The expression for the power in terms of voltage and resistance is as follows;

P=\frac{V^{2}}{R}

Here, R is the resistance.

From the above expression, it can be concluded that the power, the number of joules per second supplied by the battery is inversely proportional to the resistance. The equivalent resistance increases if the seventh resistance is connected with a battery.

If a seventh resistor is added (in series) the number of joules per second supplied by the battery decreases.

Therefore, the option (b) is correct.

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3 years ago
How can voltage be induced in a wire with the help of a magnet?
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3 0
3 years ago
Consider a satellite in a circular orbit around the Earth. If it were at an altitude equal to twice the radius of the Earth, 2RE
Elenna [48]

Answer:

v=\sqrt{\frac{gR_E}{2}}

Explanation:

Satellites experiment a force given by Newton's Gravitation Law:

F=\frac{GMm}{r^2}

where M is Earth's mass, m the satellite's mass, r the distance between their gravitational centers and G the gravitational constant.

We also know from Newton's 2nd Law that <em>F=ma, </em>so putting both together we will have:

ma=\frac{GMm}{r^2}

a=\frac{GM}{r^2}

If we are on the surface of the Earth, the acceleration would be g and r=R_E (Earth's radius):

g=\frac{GM}{R_E^2}

Which we will write as:

gR_E^2=GM

If we are on orbit the acceleration is centripetal (a=\frac{v^2}{r}), so we have:

\frac{v^2}{r}=a=\frac{GM}{r^2}=\frac{gR_E^2}{r^2}

v^2=\frac{gR_E^2}{r}

v=\sqrt{\frac{gR_E^2}{r}}

And if this orbit has a radius r=2R_E we have:

v=\sqrt{\frac{gR_E^2}{2R_E}}=\sqrt{\frac{gR_E}{2}}

3 0
2 years ago
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