Average oceanic crust thickness = (3+6)/2 = 4.5 mi
Average continental crust thickness = (20+30)/2 = 25 mi
Average Earth's crust thickness = (4.5+25)/2 = 14.75 mi
Answer:
his acceleration is 3.25
Explanation:
because you do 26÷16 which =1.625×2=3.25
Answer:
![-2.5\ \text{m/s}^2](https://tex.z-dn.net/?f=-2.5%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
![-5500\ \text{N}](https://tex.z-dn.net/?f=-5500%5C%20%5Ctext%7BN%7D)
Explanation:
m = Mass of car = 2200 kg
t = Time = 10 s
v = Final velocity = 0
u = Initial velocity = 25 m/s
a = Acceleration
From the kinematic equations of linear motion we have
![v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{10}\\\Rightarrow a=-2.5\ \text{m/s}^2](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20a%3D%5Cdfrac%7Bv-u%7D%7Bt%7D%5C%5C%5CRightarrow%20a%3D%5Cdfrac%7B0-25%7D%7B10%7D%5C%5C%5CRightarrow%20a%3D-2.5%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
Acceleration of the car is ![-2.5\ \text{m/s}^2](https://tex.z-dn.net/?f=-2.5%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
Force is given by
![F=ma\\\Rightarrow F=2200\times (-2.5)\\\Rightarrow F=-5500\ \text{N}](https://tex.z-dn.net/?f=F%3Dma%5C%5C%5CRightarrow%20F%3D2200%5Ctimes%20%28-2.5%29%5C%5C%5CRightarrow%20F%3D-5500%5C%20%5Ctext%7BN%7D)
Net force required to produce the acceleration is
.
In the cytoplasm, molecules of Glucose are broken down to smaller molecules hence releasing mechanical energy during a process called called respiration or simply 'breathing process' in Physics realm.
Refer to the diagram shown below.
Neglect wind resistance.
g = 9.8 m/s².
Quantities are measured as positive upward.
When the ball is launched, it has a vertical velocity of - 3.5 m/s and a horizontal velocity of 25 m/s.
The time, t, for the ball to reach the ground is
(-3.5 m/s)*( t s) - 0.5*(9.8 m/s²)*( t s)² = - (270 m)
-3.5t - 4.9t² = - 270
4.9t² + 3.5t - 270 = 0
t² + 0.71433t - 55.102 = 0
Solve with the quadratic formula.
t = 0.5[-0.71433 +/- √220.918] = 7.0745 s or -7.788 s.
Reject negative time.
The horizontal distance traveled is
d = (25 m/s)*(7.0745 s) = 176.8625 m
Answer:
The horizontal distance from the girl to the ball will be 176.9 m (nearest tenth)