Answer:
Fault lines
Explanation:
Earthquakes are most likely to occur near or on fault lines. A great example of this is the ring of fire, a gigantic fault line that gives catastrophic earthquakes.
The correct answer (sample response) is:
The image seems to be behind the mirror, but nothing is really there.
Include the following in your response:
The image appears to be behind the mirror.
If someone looks behind the mirror, there is no image there.
Hey there!
We are given ,
Acceleration, a = -2m/s^2
Initial velocity , u = 15m/s
Time , t = 5 seconds
We know that ,
V=u+at
Now , final speed ,
V = 15+(-2)(5)
V = 15-10
V = 5 m/s -> final speed
Hope this helps you dear :)
Have a good day <3
Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
Answer:
306 m/s
Explanation:
Law of conservation of momentum
m1v1 + m2v2 = (m1+m2)vf
m1 is the bullet's mass so it is 0.1 kg
v1 is what we're trying to solve
m2 is the target's mass so it is 5.0 kg
v2 is the targets velocity, and since it was stationary, its velocity is zero
vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s
plugging in, we get
(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)
(0.1)(v1) + 0 = 30.6
(0.1)(v1) = 30.6
v1 = 306 m/s
