Answer:
6.96 s
Explanation:
The period of a simple pendulum is given by:
![T=2 \pi \sqrt{\frac{L}{g}}](https://tex.z-dn.net/?f=T%3D2%20%5Cpi%20%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D)
where
L is the length of the pendulum and g the acceleration due to gravity.
In this problem, we have a pendulum with length L = 2.00 m, while the acceleration due to gravity is 1/6 that of the earth:
![g' = \frac{g}{6}=\frac{9.8 m/s^2}{6}=1.63 m/s^2](https://tex.z-dn.net/?f=g%27%20%3D%20%5Cfrac%7Bg%7D%7B6%7D%3D%5Cfrac%7B9.8%20m%2Fs%5E2%7D%7B6%7D%3D1.63%20m%2Fs%5E2)
So, the period of the pendulum on the moon is
![T=2 \pi \sqrt{\frac{2.00 m}{1.63 m/s^2}}=6.96 s](https://tex.z-dn.net/?f=T%3D2%20%5Cpi%20%5Csqrt%7B%5Cfrac%7B2.00%20m%7D%7B1.63%20m%2Fs%5E2%7D%7D%3D6.96%20s)
Answer:
Height of the image is 9.06 cm
Explanation:
As we know that the person is standing 250 cm from the lens
Now we know that the focal length of the lens is
f = 15 cm
now by lens formula we have
![\frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7B1%7D%7Bd_i%7D%20%2B%20%5Cfrac%7B1%7D%7Bd_o%7D)
![\frac{1}{15} = \frac{1}{d_i} + \frac{1}{250}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B15%7D%20%3D%20%5Cfrac%7B1%7D%7Bd_i%7D%20%2B%20%5Cfrac%7B1%7D%7B250%7D)
![d_i = 15.96 cm](https://tex.z-dn.net/?f=d_i%20%3D%2015.96%20cm)
now we know by the formula of magnification
![\frac{h_i}{h_o} = \frac{15.96}{250}](https://tex.z-dn.net/?f=%5Cfrac%7Bh_i%7D%7Bh_o%7D%20%3D%20%5Cfrac%7B15.96%7D%7B250%7D)
now we have
![h_o = 142 cm](https://tex.z-dn.net/?f=h_o%20%3D%20142%20cm)
![h_i = 142\times \frac{15.96}{250}](https://tex.z-dn.net/?f=h_i%20%3D%20142%5Ctimes%20%5Cfrac%7B15.96%7D%7B250%7D)
![h_i = 9.06 cm](https://tex.z-dn.net/?f=h_i%20%3D%209.06%20cm)
Answer:
as its mass and velocity will less so its momentum will be less than that of baseball
Answer:
Impulse = 322.5[kg*m/s], the answer is D
Explanation:
This method it is based on the principle of momentum and the amount of movement; and used to solve problems involving strength, mass, speed and time.
If units of the SI are used, the magnitude of the impulse of a force is expressed in N * s. however, when remembering the definition of the newton.
![N*S=(kg*m/s^{2} )*s = kg*m/s](https://tex.z-dn.net/?f=N%2AS%3D%28kg%2Am%2Fs%5E%7B2%7D%20%29%2As%20%3D%20kg%2Am%2Fs)
Now replacing the values on the following equation that express the definition of impulse
![Impulse = Force * Time\\\\Impulse = 215 * 1.5 = 322.5 [kg*m/s]](https://tex.z-dn.net/?f=Impulse%20%3D%20Force%20%2A%20Time%5C%5C%5C%5CImpulse%20%3D%20215%20%2A%201.5%20%3D%20322.5%20%5Bkg%2Am%2Fs%5D)
Answer:
![F=1.02x10^{-3} N](https://tex.z-dn.net/?f=F%3D1.02x10%5E%7B-3%7D%20N)
Explanation:
From the exercise we know:
![m=51g*\frac{1kg}{1000g}=0.051kg](https://tex.z-dn.net/?f=m%3D51g%2A%5Cfrac%7B1kg%7D%7B1000g%7D%3D0.051kg)
![v_{1}=-22m/s](https://tex.z-dn.net/?f=v_%7B1%7D%3D-22m%2Fs)
![v_{2}=14m/s](https://tex.z-dn.net/?f=v_%7B2%7D%3D14m%2Fs)
![t_{2}-t_{1}=1800s](https://tex.z-dn.net/?f=t_%7B2%7D-t_%7B1%7D%3D1800s)
So, the average acceleration is:
![a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}=\frac{(14-(-22))m/s}{1800s}=0.02m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv_%7B2%7D-v_%7B1%7D%7D%7Bt_%7B2%7D-t_%7B1%7D%7D%3D%5Cfrac%7B%2814-%28-22%29%29m%2Fs%7D%7B1800s%7D%3D0.02m%2Fs%5E2)
The average force is:
![F=m*a=(0.051kg)(0.02m/s^2)=1.02x10^{-3} N](https://tex.z-dn.net/?f=F%3Dm%2Aa%3D%280.051kg%29%280.02m%2Fs%5E2%29%3D1.02x10%5E%7B-3%7D%20N)