A hunter who is standing 220.0-m away fires his rifle. How long (in seconds) will it take the sound to travel to your ear.

Physics

1 answer:

natta225 [31]3 years ago

6 0

Answer:

t = 1.52 s

Explanation:

The speed of sound is constant and depends only on the properties of the medium in which the sound propagates, many of these properties change with temperature, in the case of propagation in air, the speed of sound

v = $331 \ \sqrt{1+ \frac{T}{273} }$

where the temperature is in degrees centigrade

v = 331 $\sqrt{1+ \frac{6}{273} }$ = 331 1.0109

v = 334.6 m / s

let's use the uniform motion relationships

v = x / t

t = v / x

let's calculate

t = 334.6 / 220.0

t = 1.52 s

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Elenna [48]

The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
$g = 9.81 \frac{m}{s {}^{2} }$
let's do it for this case:
$f_{g} = m \times g \\ f _{g} = 65kg \times 9.81 \frac{m}{s {}^{2} } = 637.65$
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
$f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} } \\ f = 325n$
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere

8 0

3 years ago

The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7m when leaving the ground at an angle of 45

zheka24 [161]

Answer:

v = 12 m/s

Long, boring, and convoluted explanation:

First, let's lay out our information.

- max height = 3.7 m

- 0 = 45°

- gravitational acceleration constant = 9.8 $\frac{m}{s^2}$

Since the puma leaves the ground at a 45 ° angle, its motion will follow a curved path as seen in many projectile motion problems, where the object is being influenced solely by the force of gravity. And because the puma leaves the ground at an angle, its initial velocity is broken down into its horizontal and vertical components. We were also told, though indirectly, that the max height is 3.7m because the puma can reach up to that height. Gravity is always given to be 9.8 $\frac{m}{s^2}$

Because we are dealing with maximum height and gravity, we have to use the vertical component of the velocity, vsin ( θ ) , and not the horizontal component, vcos ( θ ) .

Given its max height, the acceleration due to gravity, and the angle, we can now solve for the speed at which the puma leaves the ground using the following equation: vsin ( θ ) = $\sqrt{2hg}$