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:
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon)
A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by. E=hf=hcλ(energy of a photon) E = h f = h c λ (energy of a photon) , where E is the energy of a single photon and c is the speed of light.
Answer:
ok so
Explanation:
Im not sure rn but ill get back to you.
Answer:
43.68 J
Explanation:
Distance moved= 7.8 m
Force = 5.6 N
Work Done= Distance moved * Force
= 7.8 *5.6
=43.68 Joules
Answer:
Explanation:
<u>Diagonal Launch
</u>
It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.
The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is
Where vo is the magnitude of the initial velocity, 
is the angle, t is the time and g is the acceleration of gravity
The maximum height the object can reach can be computed as
There are two times where the value of y is 
when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making 
Removing and dividing by t (t different of zero)
Then we find the total flight as
We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is
