Forces on a Baseball. When a baseballis thrown or hit, the resulting motion of the ball is determined by Newton's laws of motion. ... Lift and drag are actually two components of a single aerodynamic force acting on the ball. Drag acts in a direction opposite to the motion, and lift acts perpendicular to the motion.
Answer:
the magnitude of momentum is √2≈ b
Explanation:
hope that helped
Answer:
maximum speed 56 km/h
Explanation:
To apply Newton's second law to this system we create a reference system with the horizontal x-axis and the Vertical y-axis. In this system, normal is the only force that we must decompose
sin 10 = Nx / N
cos 10 = Ny / N
Ny = N cos 10
Nx = N sin 10
Let's develop Newton's equations on each axis
X axis
We include the force of friction towards the center of the curve because the high-speed car has to get out of the curve
Nx + fr = m a
a = v2 / r
fr = mu N
N sin10 + mu N = m v² / r
N (sin10 + mu) = m v² / r
Y Axis
Ny -W = 0
N cos 10 = mg
Let's solve these two equations,
(mg / cos 10) (sin 10 + mu) = m v² / r
g (tan 10 + μ / cos 10) = v² / r
v² = r g (tan 10 + μ / cos 10)
They ask us for the maximum speed
v² = 30.0 9.8 (tan 10+ 0.65 / cos 10)
v² = 294 (0.8364)
v = √(245.9)
v = 15.68 m / s
Let's reduce this to km / h
v = 15.68 m / s (1 km / 1000m) (3600s / 1h)
v = 56.45 km / h
This is the maximum speed so you don't skid
Answer:

Explanation:
given,
radius of curve = 166 m
angle of the banked road = 11°
mass of car = 736 Kg
speed of the curve = 81 km/h
= 81 x 0.278 = 22.52 m/s
normal force acting on the tires
on tire there will be two force acting on it
first one will be force acting due to weight and the other force acting on the tire is due to centripetal force.




Answer:
(a) the observed frequency is 200 Hz
(b) the observed frequency is 188 Hz.
Explanation:
speed of the truck, Vs = 27 m/s
frequency of the truck as it approaches, Fs = 185 Hz
(a) Apply Doppler effect to determine the frequency you will hear.
As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.
![F_s = F_o [\frac{V}{V_S + V} ]](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7BV%7D%7BV_S%20%2B%20V%7D%20%5D)
Where;
Fo is the observed frequency which is the frequency you will hear.
V is speed of sound in air

(b) Apply the following formula for a moving observer and a moving source;
](https://tex.z-dn.net/?f=F_o%20%3D%20F_s%5B%5Cfrac%7BV-V_o%7D%7BV%7D%20%5D%28%5Cfrac%7BV%7D%7BV-V_S%7D%20%29)
The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.
\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz](https://tex.z-dn.net/?f=F_o%20%3D%20F_s%5B%5Cfrac%7BV-V_o%7D%7BV%7D%20%5D%28%5Cfrac%7BV%7D%7BV-V_S%7D%20%29%5C%5C%5C%5CF_o%20%3D%20185%5B%5Cfrac%7B340-22%7D%7B340%7D%20%5D%28%5Cfrac%7B340%7D%7B340-27%7D%20%29%5C%5C%5C%5CF_o%20%3D%20185%280.9353%29%281.0863%29%5C%5C%5C%5CF_o%20%3D%20188%20%5C%20Hz)