To solve this problem it is necessary to apply the kinematic equations of motion.
By definition we know that the position of a body is given by
Where
Initial position
Initial velocity
a = Acceleration
t= time
And the velocity can be expressed as,
Where,
For our case we have that there is neither initial position nor initial velocity, then
With our values we have 
, rearranging to find a,
Therefore the final velocity would be
Therefore the final velocity is 81.14m/s
Answer:
Gamma rays
Explanation:
Gamma rays is at the end of the electromagnetic spectrum, and has the highest energy. It propagates through space at 3x10^8 m/s and has the smallest wavelength and the highest frequency. It is given off by atoms of element as they undergo nuclear disintegration.
Answer:
The velocity of the plane at take off is 160 m/s.
The distance travel by the plane in that time is 3200 meter.
Explanation:
Given:
Acceleration, a = 4 m/s²
Time, t = 40 s
u = 0 i .e initial velocity
To Find:
velocity , v = ?
distance , s =?
Solution:
we have first Kinematic equation
v = u + at
∴ v = 0 + 4×40
∴ v = 160 m/s
Now by Third Kinematic equation
∴ s = 0 + 0.5 × 4× 40²
∴ s = 3200 meter
I'm pretty sure its Venus!!!
Answer:
vf = 11.2 m/s
Explanation:
m = 10 Kg
F = 2*10² N
x = 4.00 m
μ = 0.44
vi = 0 m/s
vf = ?
We can apply Newton's 2nd Law
∑ Fx = m*a (→)
F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m
⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²
then , we use the equation
vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)
⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s
