Answer:
The magnitude of gravitational force between two masses is 
.
Explanation:
Given that,
Mass of first lead ball, 
Mass of the other lead ball, 
The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m
We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :
So, the magnitude of gravitational force between two masses is . Hence, this is the required solution.
Answer:a)1.11×10^-21Nm
b) 1.16×10^-3m
Explanation:see attachment
Answer:
t = 1.4[s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.
where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 50 [kg]
v = velocity [m/s]
F = force = 200[N]
t = time = [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.
where:
m₁ = mass of the object = 50 [kg]
v₁ = velocity of the object before the impulse = 18.2 [m/s]
v₂ = velocity of the object after the impulse = 12.6 [m/s]
![(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]](https://tex.z-dn.net/?f=%2850%2A18.2%29-200%2At%3D50%2A12.6%5C%5C910-200%2At%3D630%5C%5C200%2At%3D910-630%5C%5C200%2At%3D280%5C%5Ct%3D1.4%5Bs%5D)
Answer:
0.304 m/s2
Explanation:
If the first child is pushing with a force of 69N to the right and the 2nd child is pushing with a force of 91N to the left. Then the net pushing force is 91 - 69 = 22 N to the left. Subtracted by 15N friction force then the system of interest is subjected to F = 7 N net force tot he left.
We can use Newton's 2nd law to calculate the net acceleration of the system
Answer:
4000 Hz
Explanation:
An anti-alias filter is usually added in front of the ADC to limit a certain range of input frequencies in order to avoid aliasing. This filter is usually a low pass filter that passes low frequencies but attenuates the high frequencies.
The Nyquist sampling criteria states that the sampling rate should be at least twice the maximum frequency component of the desired signal.
Sampling rate = 2(max input frequency)
From the relation we can find out the cut-off frequency for the anti-aliasing filter.
max input frequency = sampling rate/2
max input frequency = 8100/2 = 4050 Hz
Therefore, 4000 Hz would be an appropriate cut-off frequency for the anti-aliasing filter.
