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3 years ago

14

While filming an intense action sequence for the next James Bond movie, a controlled explosion detonates 1.1 km away from the ac

tors. If the speed of sound through solid rock is 3000 m/s on average, the actors will feel the explosion before they hear it. How much time will pass between when they feel the explosion and when they hear it?

1 answer:

kolezko [41]3 years ago

4 0

Answer:

The time that will pass between the feeling and hearing the explosion is 2,86 secs

Explanation:

First, let's calculate the time that the wave takes to travel until the actors feel the explosion:

$1,1 km*\frac{1.000 mts}{1 km} *\frac{sec}{3.000 mts} = 0,37 secs$

Now, the time that pass while the actors hear the sound is:

<em>(Remember that the sound speed in the air is 340 m/s on average)</em>

$1,1 km * \frac{1.000 mts}{1 km} * \frac{sec}{340 mts} = 3,23 secs$

So, the time between the feeling and hearing is 3,23 - 0,37 = 2,86 secs

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Answer: Gamma-rays have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum. These waves are generated by radioactive atoms and in nuclear explosions. Gamma-rays can kill living cells, a fact which medicine uses to its advantage, using gamma-rays to kill cancerous cells.

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1 year ago

In the Rolling Plains ecoregion of Texas, some parts of rivers are in deep canyons and others have bottoms that are close to the

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Explanation:

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3 years ago

Read 2 more answers

A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp

maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance $D$ , which is divided into two segments $d=10 miles$ :

$D=d+d=2d$ (1)

Where $d=10mi \frac{1609.34 m}{1 mi}=16093.4 m$

On the other hand, we know speed is defined as:

$S=\frac{d}{t}$ (2)

Where $t$ is the time, which can be isolated from (2):

$t=\frac{d}{S}$ (3)

Now, for the first segment $d=16093.4 m$ the car has a speed $S_{1}=22m/s$ , using equation (3):

$t_{1}=\frac{d}{S_{1}}$ (4)

$t_{1}=\frac{16093.4 m}{22m/s}$ (5)

$t_{1}=731.518 s$ (6) This is the time it takes to travel the first segment

For the second segment $d=16093.4 m$ the car has a speed $S_{1}=30m/s$ , hence:

$t_{2}=\frac{d}{S_{2}}$ (7)

$t_{2}=\frac{16093.4 m}{30m/s}$ (8)

$t_{2}=536.44 s$ (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed $S_{ave}$ :

$S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}}$ (10)

$S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s}$ (11)

Finally:

$S_{ave}=25.38 m/s$

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2 years ago

If the mass of a

mixer [17]

Answer:

The final acceleration becomes (1/3) of the initial acceleration.

Explanation:

The second law of motion gives the relationship between the net force, mass and the acceleration of an object. It is given by :

$F=ma$

m = mass

a = acceleration

According to given condition, if the mass of a sliding block is tripled while a constant net force is applied. We need to find how much does the acceleration decrease.

$a=\dfrac{F}{m}$

Let a' is the final acceleration,

$a'=\dfrac{F}{m'}$

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$a'=\dfrac{F}{3m}$

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$a'=\dfrac{1}{3}\times a$

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2 years ago

In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separatio

Keith_Richards [23]

Answer:

b. 0.25cm

Explanation:

You can solve this question by using the formula for the position of the fringes:

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d: separation of the slits 1mm=1*10^{-3}m

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$\Delta y=\frac{(m+1)(\lambda D)}{d}-\frac{m\lambda D }{d}=\frac{\lambda D}{d}\\\\\Delta y=\frac{(500*10^{-9}nm)(5m)}{1*10^{-3}m}=2.5*10^{-3}m=0.25cm$

hence, the aswer is 0.25cm

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2 years ago