Answer:
It is possible by increasing the speed of the tennis ball by a factor of (Mass of the tennis ball)/(Mass of the basketball)
Explanation:
The momentum of a body = The bod's mass × The body's velocity
Therefore, the momentum of a given mass of an object, such as a tennis ball can be increased by increasing the velocity or speed of the object. Whereby the speed of the ball, v₁, is increased such that the momentum of the basketball and the tennis ball will be the same, is given by the following equation
Mass of the basketball × v₂ = Mass of the tennis ball × v₁
Therefore, v₁/v₂ = (Mass of the tennis ball)/(Mass of the basketball)
Answer:
The observed frequency by the pedestrian is 424 Hz.
Explanation:
Given;
frequency of the source, Fs = 400 Hz
speed of the car as it approaches the stationary observer, Vs = 20 m/s
Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.
The observed frequency is calculated as;
![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%20%5C%5C%5C%5C)
where;
F₀ is the observed frequency
v is the speed of sound in air = 340 m/s
![F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\](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%20%5C%5C%5C%5C400%20%3D%20F_o%20%5B%5Cfrac%7B340%7D%7B20%20%2B%20340%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%280.9444%29%20%5C%5C%5C%5CF_o%20%3D%20%5Cfrac%7B400%7D%7B0.9444%7D%20%5C%5C%5C%5CF_o%20%3D%20423.55%20%5C%20Hz%20%5C%5C)
F₀ ≅ 424 Hz.
Therefore, the observed frequency by the pedestrian is 424 Hz.
Force = -kx
80N=0.15m * -k
K=-80/0.15=533.333. Spring constant
Energy=1/2kx^2
1/2*(-80/0.15)*80^2=Energy
Answer:
(a) T= 38.4 N
(b) m= 26.67 kg
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Kinematics
d= v₀t+ (1/2)*a*t² (Formula 2)
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
v₀=0, d=18 m , t=5 s
We apply the formula 2 to calculate the accelerations of the blocks:
d= v₀t+ (1/2)*a*t²
18= 0+ (1/2)*a*(5)²
a= (2*18) / ( 25) = 1.44 m/s²
to the right
We apply Newton's second law to the block A
∑Fx = m*ax
60-T = 15*1.44
60 - 15*1.44 = T
T = 38.4 N
We apply Newton's second law to the block B
∑Fx = m*ax
T = m*ax
38.4 = m*1.44
m= (38.4) / (1.44)
m = 26.67 kg
