Answer:
I will but can you just wait for some minutes cus I am in a hurry now.
sorry that pic is a little blurry
Answer:
Current = dQ/dt
or I = dQ/dt
Where I represents current.
Which is the rate of flow of charge.
Q=4 + 2t + t²
dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.
At time t=2sec
dQ/dt = I = 2 + 2t
= 2 + 2(2)
=2 + 4
= 6A.
Answer:
103.5 meters
Explanation:
Given that a stunt person has to jump from a bridge and land on a boat in the water 22.5 m below. The boat is cruising at a constant velocity of 48.3 m/s towards the bridge. The stunt person will jump up at 6.45 m/s as they leave the bridge.
The time the person will jump to a certain spot under the bridge can be calculated by using the formula below:
h = Ut + 1/2gt^2
since the person will fall under gravity, g = 9.8 m/s^2
Also, let assume that the person jump from rest, then, U = 0
Substitute h, U and g into the formula above
22.5 = 1/2 * 9.8 * t^2
22.5 = 4.9t^2
22.5 = 4.9t^2
t^2 = 22.5/4.9
t^2 = 4.59
t = 
t = 2.143 seconds
From definition of speed,
speed = distance /time
Given that the boat is cruising at a constant velocity of 48.3 m/s towards the bridge, substitute the speed and the time to get the distance.
48.3 = distance / 2.143
distance = 48.3 * 2.143
distance = 103.5 m
Therefore, the boat should be 103.5m away from the bridge at the moment the stunt person jumps?
Answer:
It's energy will double.
Explanation:
This is because energy, E, is related to frequency, f, by:
E = hf
Where h = Planck's constant
So, double frequency will be 2f
=> E(2f) = 2hf = 2E.
Hence, energy is doubled.
To solve this problem we will apply the momentum conservation theorem, that is, the initial momentum of the bodies must be the same final momentum of the bodies. The value that will be obtained will be a vector value of the final speed of which the magnitude will be found later. Our values are given as,




Using conservation of momentum,


Solving for 

Using the properties of vectors to find the magnitude we have,


Therefore the magnitude of the velocity of the wreckage of the two cars immediately after the collision is 12.4135m/s