The answer is:
<span> 4. 1 and 2 only.</span>
They are built to be airtight so thsy sir can't get in or out, and then, as they climb up to where the air is thin, they use an air pump to pimp air into the inside.
Just like a party balloon.
Answer:
C
Explanation:
Spoon decrease, water increase
<span>51 degrees.
Since we're ignoring friction, we have to have a banking angle such that the normal force is exactly perpendicular to the banked curve. Since this problem says "ignore friction", if the bank angle is too shallow, the bobsled would slide outwards if the banking angle is too shallow and would fall inwards if the banking angle is too steep. So we have to exactly match the calculated centripetal acceleration.
The equation for centripetal acceleration is:
F = mv^2/r
I'll assume a mass of 1 kg to keep the math simple. Any mass could be used and the direction vectors would be the same except their magnitude would differ. So
F = 1 kg * (35 m/s)^2/100 m
F = 1225 kg*m^2/s^2 / 100 m
F = 12.25 kg*m/s^2
The local gravitational acceleration is 9.8 m/s^2, so the sum of those vectors will have a length of sqrt(12.25^2 + 9.8^2) and an angle of atan(9.8/12.25) below the horizon. The magnitude of the vector doesn't matter, merely the angle which is:
atan(9.8/12.25) = atan(0.8) = 38.65980825 degrees.
The banking angle needs to be perpendicular to the force vectors. So
90 - 38.65980825 = 51.34019175 degrees.
Rounding to 2 significant figures gives a bank angle of 51 degrees.</span>
Answer:
d = 771.3m
Explanation:
Let's first calculate the time of flight:
where Voy=0. Solving for t:

Now we calculate the horizontal displacement, wihch is the distance from the target to drop the package:
Xf = d = Vox*t
d = 180*4.285 = 771.3m