Answer:
0.775 m
Explanation:
As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):
![U=K\\frac{1}{2}kx^2 = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=U%3DK%5C%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where we have
is the spring constant of the bumper
x is the maximum compression of the bumper
is the mass of the car
is the speed of the car
Solving for x, we find the maximum compression of the spring:
![x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B%5Cfrac%7Bmv%5E2%7D%7Bk%7D%7D%3D%5Csqrt%7B%5Cfrac%7B%2830%5Ccdot%2010%5E4%20kg%29%282.0%20m%2Fs%29%5E2%7D%7B2%5Ccdot%2010%5E6%20N%2Fm%7D%7D%3D0.775%20m)
If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Given the data in the question;
- Length of the massless beam;
![L = 4.00m](https://tex.z-dn.net/?f=L%20%3D%204.00m)
- Distance of support from the left end;
![x = 3.00m](https://tex.z-dn.net/?f=x%20%3D%203.00m)
- First mass;
![m1 = 31.3 kg](https://tex.z-dn.net/?f=m1%20%3D%2031.3%20kg)
- Distance of beam from the left end( m₁ is attached to );
![x_1 = ?](https://tex.z-dn.net/?f=x_1%20%3D%20%3F)
- Second mass;
![m_2 = 61.7 kg](https://tex.z-dn.net/?f=m_2%20%3D%2061.7%20kg)
- Distance of beam from the right of the support( m₂ is attached to );
![x_1 = 0.273m](https://tex.z-dn.net/?f=x_1%20%3D%200.273m)
Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.
Hence, ![m_1g( x-x_1) = m_2gx_2](https://tex.z-dn.net/?f=m_1g%28%20x-x_1%29%20%3D%20m_2gx_2)
we divide both sides by ![g](https://tex.z-dn.net/?f=g)
![m_1( x-x_1) = m_2x_2](https://tex.z-dn.net/?f=m_1%28%20x-x_1%29%20%3D%20m_2x_2)
Next, we make
, the subject of the formula
![x_1 = x - [ \frac{m_2x_2}{m_1} ]](https://tex.z-dn.net/?f=x_1%20%3D%20x%20-%20%5B%20%5Cfrac%7Bm_2x_2%7D%7Bm_1%7D%20%5D)
We substitute in our given values
![x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]](https://tex.z-dn.net/?f=x_1%20%3D%203.00m%20-%20%5B%20%5Cfrac%7B61.7kg%5C%20%2A%20%5C%200.273m%7D%7B31.3kg%7D%20%5D)
![x_1 = 3.00m - 0.538m](https://tex.z-dn.net/?f=x_1%20%3D%203.00m%20-%200.538m)
![x_1 = 2.46m](https://tex.z-dn.net/?f=x_1%20%3D%202.46m)
Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Learn more; brainly.com/question/3882839
Answer:
The answer is c. 11.42 Ohm
Explanation:
The conductor's resistance is calculated by the formula in the figure.
So, you have to replace the given values into the formula.
Resistance of a conductor is equal to the product of rho by the lengh of the conductor divided the cross-sectional area of the conductor.
![R= 4 ohm.m . (2m/o.7m^{2} )\\R=11.42ohm](https://tex.z-dn.net/?f=R%3D%204%20ohm.m%20.%20%282m%2Fo.7m%5E%7B2%7D%20%29%5C%5CR%3D11.42ohm)
I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.
If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.
For the wave you described . . .
Frequency = (speed) / (wavelength)
= (6 m/s) / (12 m)
= 0.5 / sec
= 0.5 Hz .
(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)
Hi! The answer is ‘B’! Because the nucleus is found at the center and contains protons (positive charge) and neutrons (no charge)