Answer:
One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.
Explanation:
This hypothesis is based on the fact that the speed of sound in air is v = 343 m / s with a small variation with temperature.
The speed of sound in solid soil is an average of the speed of its constituent media, giving values between
wood 3900 m / s
concrete 4000 m / s
fabrics 1540 m / s
earth 5000 m / s wave S
ground 7000 m / s P wave
we can see that the speed on solid earth is an order of magnitude greater than in air.
One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.
From the initial information, the wave going through the ground should arrive first.
Answer:
it is 1250 because it moves 5000 in four hours then it means that it is moving at 1250 an hour
Explanation:
Answer:
λ = 6.97 *10⁻⁶ C/m
Explanation:
Conceptual Analysis:
The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:
E= 2k*(λ/r) Formula (1)
Where:
E: electric field .( N/C)
k: Coulomb electric constant. (N*m²/C²)
λ: linear charge density. (C/m)
r : distance from the charge line to the surface where E calculates (m)
Known data
E= 6.6 x 10⁴ N/C
r = 1.9 m
k= 8.99 *10⁹ N*m²/C²
Problem development
We replace data in the formula (1):
E= 2k*(λ/r)
6.6*10⁴ = 2*(8.99 *10⁹)*(λ/1.9)
(6.6*10⁴)*(1.9) = 2*(8.99 *10⁹) *λ
(12.54* 10⁴) / (17.98*10⁹) = λ
λ = 6.97 *10⁻⁶ C/m
Answer:
![\frac{t_1}{t_2} = \frac{sin\theta_1}{sin\theta_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bt_1%7D%7Bt_2%7D%20%3D%20%5Cfrac%7Bsin%5Ctheta_1%7D%7Bsin%5Ctheta_2%7D)
Explanation:
The vertical component of the initial velocities are
![v_v = v_0sin\theta](https://tex.z-dn.net/?f=v_v%20%3D%20v_0sin%5Ctheta)
If we ignore air resistance, and let g = -9.81 m/s2. The the time it takes for the projectiles to travel, vertically speaking, can be calculated in the following motion equation
![v_vt - gt^2/2 = s = 0](https://tex.z-dn.net/?f=v_vt%20-%20gt%5E2%2F2%20%3D%20s%20%3D%200)
![t(v_v - gt/2) = 0](https://tex.z-dn.net/?f=t%28v_v%20-%20gt%2F2%29%20%3D%200)
![v_v - gt/2 = 0](https://tex.z-dn.net/?f=v_v%20-%20gt%2F2%20%3D%200)
![t = 2v_v/g = 2v_0sin\theta/g](https://tex.z-dn.net/?f=t%20%3D%202v_v%2Fg%20%3D%202v_0sin%5Ctheta%2Fg)
So the ratio of the times of the flights is
![t_1 / t_2 = \frac{2v_0sin\theta_1/g}{2v_0sin\theta_2/g} = \frac{sin\theta_1}{sin\theta_2}](https://tex.z-dn.net/?f=t_1%20%2F%20t_2%20%3D%20%5Cfrac%7B2v_0sin%5Ctheta_1%2Fg%7D%7B2v_0sin%5Ctheta_2%2Fg%7D%20%3D%20%5Cfrac%7Bsin%5Ctheta_1%7D%7Bsin%5Ctheta_2%7D)