Answer: The distance is 723.4km
Explanation:
The velocity of the transverse waves is 8.9km/s
The velocity of the longitudinal wave is 5.1 km/s
The transverse one reaches 68 seconds before the longitudinal.
if the distance is X, we know that:
X/(9.8km/s) = T1
X/(5.1km/s) = T2
T2 = T1 + 68s
Where T1 and T2 are the time that each wave needs to reach the sesmograph.
We replace the third equation into the second and get:
X/(9.8km/s) = T1
X/(5.1km/s) = T1 + 68s
Now, we can replace T1 from the first equation into the second one:
X/(5.1km/s) = X/(9.8km/s) + 68s
Now we can solve it for X and find the distance.
X/(5.1km/s) - X/(9.8km/s) = 68s
X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s
X = 68s/0.094s/km = 723.4 km
Answer:
Option (A) , (b) and (d) are correct option
Explanation:
According to Coulomb's law electric force between two charges is given by

From the relation we can say that force is directly proportional to magnitude of charges and inversely proportional to distance between them '
So if we increase the distance then force will decrease
Increase if any of the charge get larger
If force is attractive then both the charge will be of different sign and is force is repulsive then both the charges of same sign
From above conclusion we can say that (a), (b) and (d) are correct option

Jake was playing making a paper airplane, after making he kept it on the table and went to have food.
Suddenly his brother saw the plane and threw it in the air, The airplane kept flying for about 3m/s and it hit his mother and due to the force the plane stopped.
{kept in on the table>rest
brother threw the plane>moving
it hit his mother>force that stopped it}
(mark me brainliest if you're satisfied with my answer)

Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
n₁sinθ₁ = n₂sinθ₂
y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and
θ₁ = 14.511 °
n₁ = 1.0003 = refractive index of air
n₂ = 1.33 = refractive index of water
Therefore sinθ₂ =
=
= 0.1885 and θ₂ = 10.86 °
Since the water depth is 4.0 m we have tanθ₂ =
or x₂ =
=
= 20.845 m
d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.