#1). The horizontal speed doesn't change.
The vertical speed is accelerated by gravity.
a). Gravity increases the vertical speed by 9.8m/s every sec.
After 5 sec, the car is falling (5x9.8) = <em>49 m/s</em> vertically.
b). Horizontal: 49m/s ! Wow ! Almost 110 mph. No wonder he went off the cliff.
After 5 seconds, it's still <em>49 m/s</em>.
c). After 5 sec, the horizontal speed and vertical-down speed are both 49 m/s.
The combination results in a velocity that points 45 degrees down from horizontal,
and its magnitude is
square root of (49² + 49²) = 49 √2 = about <em>69.3 m/s</em> .
Answer:
The gate will open if the height of water is equal to or more than 0.337m.
Explanation:
From the diagram attached, (as seen from the reference question found on google)
The forces are given as
Force on OA

Here
- ρ is the density of water.
- g is the gravitational acceleration constant
is the equivalent height given as

is the area of the OA part of the door which is calculated as follows:
The Force is given as
![F_1=0.6\rho g[h+0.3]](https://tex.z-dn.net/?f=F_1%3D0.6%5Crho%20g%5Bh%2B0.3%5D)
Force on OB

Here
- ρ is the density of water.
- g is the gravitational acceleration constant
is the equivalent height given as

is the area of the OB part of the door which is calculated as follows:

The Force is given as
![F_2=0.4\rho g[h+0.8]](https://tex.z-dn.net/?f=F_2%3D0.4%5Crho%20g%5Bh%2B0.8%5D)
Now the moment arms are given as
![\bar{y}_a=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_a=h+0.3+\frac{\frac{1}{12}\times 0.6^3 \times 1}{0.6 \times[h+0.3]}\\\bar{y}_a=h+0.3+\frac{0.03}{h+0.3}](https://tex.z-dn.net/?f=%5Cbar%7By%7D_a%3D%5Cbar%7Bh%7D%2B%5Cfrac%7BI%7D%7BA%5Cbar%7Bh%7D%7D%5C%5C%5Cbar%7By%7D_a%3Dh%2B0.3%2B%5Cfrac%7B%5Cfrac%7B1%7D%7B12%7D%5Ctimes%200.6%5E3%20%5Ctimes%201%7D%7B0.6%20%5Ctimes%5Bh%2B0.3%5D%7D%5C%5C%5Cbar%7By%7D_a%3Dh%2B0.3%2B%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D)
![\bar{y}_b=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_b=h+0.8+\frac{\frac{1}{12}\times 0.4^3 \times 1}{0.4 \times[h+0.8]}\\\bar{y}_b=h+0.8+\frac{0.0133}{h+0.8}](https://tex.z-dn.net/?f=%5Cbar%7By%7D_b%3D%5Cbar%7Bh%7D%2B%5Cfrac%7BI%7D%7BA%5Cbar%7Bh%7D%7D%5C%5C%5Cbar%7By%7D_b%3Dh%2B0.8%2B%5Cfrac%7B%5Cfrac%7B1%7D%7B12%7D%5Ctimes%200.4%5E3%20%5Ctimes%201%7D%7B0.4%20%5Ctimes%5Bh%2B0.8%5D%7D%5C%5C%5Cbar%7By%7D_b%3Dh%2B0.8%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D)
Taking moment about the point O as zero
=0.4\rho g[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.6[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.18h -0.054-0.018=0.08h+0.064+0.00533\\h=0.337 m](https://tex.z-dn.net/?f=F_1%28h%2B0.6-%5Cbar%7By%7D_a%29%3DF_2%28%5Cbar%7By%7D_b-h%2B0.6%29%5C%5CF_1%28h%2B0.6-h-0.3-%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D%29%3DF_2%28h%2B0.8%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D-h-0.6%29%5C%5CF_1%280.3-%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D%29%3DF_2%280.2%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D%29%5C%5C0.6%5Crho%20g%5Bh%2B0.3%5D%280.3-%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D%29%3D0.4%5Crho%20g%5Bh%2B0.8%5D%280.2%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D%29%5C%5C0.6%5Bh%2B0.3%5D%280.3-%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D%29%3D0.4%5Bh%2B0.8%5D%280.2%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D%29%5C%5C0.18h%20-0.054-0.018%3D0.08h%2B0.064%2B0.00533%5C%5Ch%3D0.337%20m)
So the gate will open if the height of water is equal to or more than 0.337m.
To reduce the internal energy U of an ideal gas you need the gas to lose heat and/or do work.
Answer:
(A⃗ ×B⃗ )⋅C⃗ = 69.868
Explanation:
We simplify the cross product first, thereafter the solution of the cross product is now simplified with the dot product as shown in the step by step calculation in the attachment
This study of science is called Astronomy; the study of the night sky, planets, galaxies, and other celestial objects