Answer:
(1) 0.333 Hz
(2) 4 sec
(3) 2 sec, 0.5 Hz
Explanation:
(1) We have given time period of pendulum is 3 sec
So T = 3 sec
Frequency will be equal to ![f=\frac{1}{T}=\frac{1}{3}=0.333Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7BT%7D%3D%5Cfrac%7B1%7D%7B3%7D%3D0.333Hz)
(2) Frequency of the pendulum is given f = 0.25 Hz
Time period is equal to ![T=\frac{1}{f}=\frac{1}{0.25}=4sec](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7B0.25%7D%3D4sec)
(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds
So time taken to complete 1 back and forth swings = ![=\frac{20}{10}=2sec](https://tex.z-dn.net/?f=%3D%5Cfrac%7B20%7D%7B10%7D%3D2sec)
So time period T = 2 sec
Frequency will be equal to ![f=\frac{1}{T}=\frac{1}{2}=0.5Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7BT%7D%3D%5Cfrac%7B1%7D%7B2%7D%3D0.5Hz)
Answer:
The average acceleration of the ball during the collision with the wall is ![a=2,800m/s^{2}](https://tex.z-dn.net/?f=a%3D2%2C800m%2Fs%5E%7B2%7D)
Explanation:
<u>Known Data</u>
We will asume initial speed has a negative direction,
, final speed has a positive direction,
,
and mass
.
<u>Initial momentum</u>
![p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s](https://tex.z-dn.net/?f=p_%7Bi%7D%3Dmv_%7Bi%7D%3D%28-30m%2Fs%29%28m_%7Bb%7D%29%3D-30m_%7Bb%7D%5C%20m%2Fs)
<u>final momentum</u>
![p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s](https://tex.z-dn.net/?f=p_%7Bf%7D%3Dmv_%7Bf%7D%3D%2826m%2Fs%29%28m_%7Bb%7D%29%3D26m_%7Bb%7D%5C%20m%2Fs)
<u>Impulse</u>
![I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s](https://tex.z-dn.net/?f=I%3D%5CDelta%20p%3Dp_%7Bf%7D-p_%7Bi%7D%3D26m_%7Bb%7D%5C%20m%2Fs-%28-30m_%7Bb%7D%5C%20m%2Fs%29%3D56m_%7Bb%7D%5C%20m%2Fs)
<u>Average Force</u>
![F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B%5CDelta%20p%7D%7B%5CDelta%20t%7D%20%3D%5Cfrac%7B56m_%7Bb%7D%5C%20m%2Fs%7D%7B0.020s%7D%20%3D2800m_%7Bb%7D%20%5C%20m%2Fs%5E%7B2%7D)
<u>Average acceleration</u>
, so
.
Therefore, ![a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B2800m_%7Bb%7D%20%5C%20m%2Fs%5E%7B2%7D%7D%7Bm_%7Bb%7D%7D%20%3D2800m%2Fs%5E%7B2%7D)
The centripetal acceleration a is 4.32
10^-4 m/s^2.
<u>Explanation:</u>
The speed is constant and computing the speed from the distance and time for one full lap.
Given, distance = 400 mm = 0.4 m, Time = 100 s.
Computing the v = 0.4 m / 100 s
v = 4
10^-3 m/s.
radius of the circular end r = 37 mm = 0.037 m.
centripetal acceleration a = v^2 / r
= (4
10^-3)^2 / 0.037
a = 4.32
10^-4 m/s^2.
C is correct. The work-force relation is given by W=F·d, where F is force vector, and d is the displacement vector. The dot is the dot product, which is a measure of how parallel the two vectors are. It can be restated as the product of two vector magnitudes times the cosine of the angle between them. Therefore work is a scalar, not a vector, since the dot product returns a scalar.
Answer:
P= 390 W
Explanation:
In physics, power is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time. Work is a force F applied over a distance x. Matemathicaly it means
P = dW/dt ≈ d(F * x)/dt = xdF/dt + Fdx/dt. If force is constant dF/dt=0 so P=F dx/dt = P*v, where v is velocity, the rate of distance per unit time.
We have force and velocity. Newton is unit of Kg*m/s2, hence
P= 150 kgm/s2 * 2.6 m/s =390 Kgm2/s3 = 390 W, where W is Watts and is an unit of power