Answer:
300.16m/s
Explanation:
An artillery shell is fired at an angle to the horizontal. Its initial velocity has a vertical component of 150 meters per second and a horizontal component of 260 meters per second. What is the magnitude of the initial velocity of the shell?
note velocity is the change in displacement to time
to find the magnitude of the initial speed ,
we will find the square of the vertical component plus the square of the horizontal component. then we look for the square root
U=
U=
U=90,100^2
U=300.16m/s
means that it covers 300.16m in 1 seconds
Answer:
55.8W
Explanation:
P= V^2/R
R= V^2/P
For series connection
Req= R1+ R2= V^2/310 + V^2/180
R=V^2/P= V^2/310 + V^2/180
But V^2 will cancel out
P= 1/(1/310 + 1/180)
P= 55.8W
Answer:
No..
Explanation:
As the bird releases the drop there is no internal force which will drive it into circular path but it will fall on tangent of the arc at the point of release because it has a tangential velocity same as bird. Path will be parabola in vertical plane.
As the person is on circular arc constantly moving it will never meet that drop.
The complete queston is The amount of a radioactive element A at time t is given by the formula
A(t) = A₀e^kt
Answer: A(t) =N e^( -1.2 X 10^-4t)
Explanation:
Given
Half life = 5730 years.
A(t) =A₀e ^kt
such that
A₀/ 2 =A₀e ^kt
Dividing both sides by A₀
1/2 = e ^kt
1/2 = e ^k(5730)
1/2 = e^5730K
In 1/2 = 5730K
k = 1n1/2 / 5730
k = 1n0.5 / 5730
K= -0.00012 = 1.2 X 10^-4
So that expressing N in terms of t, we have
A(t) =A₀e ^kt
A₀ = N
A(t) =N e^ -1.2 X 10^-4t
