Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
Answer:
gas is dioatomic
T_f = 330.0 K

Explanation:
Part 1
below equation is used to determine the type Gas by determining
value

where V_i and V_f is initial and final volume respectively
and P_i and P_f are initial and final pressure


\gamma = 1.38
therefore gas is dioatomic
Part 2
final temperature in adiabatic process is given as
](https://tex.z-dn.net/?f=T_f%20%3D%20T_i%2A%5B%5Cfrac%7Bv_i%7D%7BV_f%7D%5D%28%5E%5Cgamma-1%29)
substituing value to get final temperature
![T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}](https://tex.z-dn.net/?f=T_f%20%3D%20260%2A%5B%5Cfrac%7B151%7D%7B80.6%7D%5D%5E%20%7B%281.38-1%29%7D)
T_f = 330.0 K
Part 3
determine number of moles by using following formula



Answer:because the parachute is built so the wind will push up on it make the man/woman glide or fall slowly will a stone which has a lot of desentity
falls through the air faster do to is weight and shape .
Explanation:
Answer:
the question is incomplete, the complete question is
"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"
2.6mA
Explanation:
we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.
using the formula be low,

where B is the magnitude of the field and A is the area of the circular coil.
First, let determine the area using
where r is the radius of 5cm or 0.05m

since we no that the angle is at
we determine the magnitude of the magnetic filed


the Magnitude of the voltage is 0.000532V
Next we determine the current using ohm's law

