QC Suppose you hear a clap of thunder 16.2s after seeing the associated lightning strike. The speed of light in air is 3.00× 10⁸
1 answer:
If you hear a clap of thunder in a time of 16.2s after seeing the associated lightning strike, you are: 5508 m far from the lightning strike
To solve this problem we must consider that the speed of light is greater than the speed of sound, therefore to calculate the distance we must use the speed of sound (340 m/s).
The formula and procedure we will use to solve this exercise is:
x = v * t
Where:
- x = distance
- t = time
- v = velocity
Information about the problem:
- v(sound) = 340 m/s
- t = 16.2 s
- x=?
Applying the distance formula we have that:
x = v * t
x= 340 m/s * 16.2 s
x = 5508 m
<h3>What is velocity?</h3>It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).
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Answer:
t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s
Explanation:
Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso
y = y₀ + t – ½ g t²
en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)
0 = y₀ – ½ g t²
t= √ (2 y₀/g)
calculemos
t= √ ( 2 12 / 9,8)
t= 1,56 s
El alcance del proyectil es la distancia horizontal recorrida
x = v₀ₓ t
x = 80 1,56
x= 124,8 m
La velocidad de impacto cuando toca el suelo
vx = v₀ₓ = 80 ms
= v_{oy} – gt
v_{y} = - 9,8 1,56
v_{y} = - 15,288 m/s
la velocidad es
v = (80 i^ - 15,288 j ) m/s
Traducttion
This is a projectile launching exercise, let's start by finding the time it takes to reach the ground
y = y₀ + v_{oy} t - ½ g t²
in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( = 0)
0 =y₀I - ½ g t²
t = √ (2y₀ / g)
let's calculate
t = √ (2 12 / 9.8)
t = 1.56 s
Projectile range is the horizontal distance traveled
x = v₀ₓ t
x = 80 1.56
x = 124.8 m
Impact speed when it hits the ground
vₓ = v₀ₓ = 80 ms
v_{y} = v_{oy} - gt
v_{y} = - 9.8 1.56
v_{y} = - 15,288 m / s
the speed is
v = (80 i ^ - 15,288 j) m / s