Ask question

Ask question

All categories

2 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]2 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

You might be interested in

If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the fr

Dmitriy789 [7]

Answer:

Frequency will be equal to 5.20 kHz

So option (c) will be correct answer

Explanation:

We have given value of capacitance $C=8.5nF=8.5\times 10^{-9}f$

Potential difference across capacitor V = 12 volt

Current through capacitor $i=3.33mA=3.33\times 10^{-3}A$

Capacitive reactance will be equal to $X_c=\frac{V}{i}=\frac{12}{3.33\times 10^{-3}A}=3603.60ohm$

Capacitive reactance is equal to $X_c=\frac{1}{\omega C}$

$3603.60=\frac{1}{\omega\times 8.5\times 10^{-9}}$

$\omega =32647.091rad/sec$

$2\pi f=32647.091$

$f=5198.98Hz$

f = 5.20 kHz

So frequency will be equal to 5.20 kHz

So option (c) will be correct answer

3 0

2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago

Three chairs were rolling across a flat office floor. The blue and red chairs are the same mass, and they have more mass than th

Scilla [17]

The answer is the second one of it that

5 0

2 years ago

Read 2 more answers

The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the

satela [25.4K]

The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
$I= \frac{P}{A}$ (1)
where
P is the power
A is the area

In our problem, the intensity is $I=6.0 \cdot 10^{-10} W/m^2$ . At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
$A=4 \pi r^2 = 4 \pi (6.0 m)^2 = 452.2 m^2$

And so if we re-arrange (1) we find the power emitted by the source:
$P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W$

3 0

2 years ago

Which graph represents the motion of an object traveling with a positive velocity and a negative acceleration?

nignag [31]

Answer
graph 2

Explanation
Velocity is the rate of change of displacement.
Velocity = (change in displacement)/(change is time)
In graph 2, the gradient of the curve is positive, indicating that the velocity is positive.
Acceleration is the rate of change of velocity.
The steepness of the slope of the same graph (graph 2) is decreasing. This shows that the velocity is decreasing hence a negative acceleration.

5 0

2 years ago

Read 2 more answers