Answer:
a)-2m/s^2
b)27.2m/s
Explanation:
Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)
W=455N=weight
W=mg
W=455N=weight
The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration
for point b we use the equations of motion with constant acceleration to find the velocity
Where
Vf = final speed
Vo = Initial speed
=0
A = acceleration
=2m/s
X = displacement
=6.8m
Solving
Explanation:
the object has constant velocity for 2 seconds and it get a constant accelration (2ms-2)
Answer:
true
Explanation:
I did this unit for science
To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as
Where,
F= Force
t= time
At the same time the moment can be described as a function of mass and velocity, that is
Where,
m = mass
v = Velocity
From equilibrium the impulse is equal to the momentum, therefore
PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be
Therefore the magnitude of the person's impulse is 1125Kg.m/s
PART B) From the equation obtained previously we have that the Force would be:
Therefore the magnitude of the average force the airbag exerts on the person is 45000N
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.
