Answer:
V = 2.87 m/s
Explanation:
The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.
Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:
Centripetal acceleration = V^2 / r
where r is the distance of water from the pivot or shoulder.
For our case, r will be 0.65 + 0.19 = 0.84 m
and solving the above equation we get:
9.81 = V^2 / 0.84
V^2 = 8.2404
V = 2.87 m/s
Bro the md that she lost was the md boneless burger
Momentum = (mv).
<span>(2110 x 24) = 50,640kg/m/sec. truck momentum. </span>
<span>Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec</span>
Answer:
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Read more on Brainly.com - brainly.com/question/14295948#readmore
Explanation:
Answer:
![E = 1.85*10^{12}\frac{N}{C}](https://tex.z-dn.net/?f=%20E%20%3D%201.85%2A10%5E%7B12%7D%5Cfrac%7BN%7D%7BC%7D%20)
Explanation:
Hi!
The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.
The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:
![E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%20%5Cepsilon_0%20r%7D%5C%5C%5Clambda%3D%5Ctext%7Bcharge%20per%20unit%20length%7D%3D%5Cfrac%7B4.95%20%5Cmu%20C%7D%7B2%20m%7D%20%3D%202.475%20%5Cfrac%7BC%7D%7Bm%7D%5C%5Cr%3D%5Ctext%7Bperpendicular%20distance%20to%20wire%7D%5C%5C%5Cepsilon_0%3D8.85%2A10%5E%7B-12%7D%5Cfrac%7BC%5E2%7D%7BNm%5E2%7D)
Then the electric field at the point of interest is estimated as:
![E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%5C22.475%7D%7B2%5Cpi%2A%28%208.85%2A10%5E%7B-12%7D%29%2A%282.4%2A10%5E%7B-2%7D%29%7D%5Cfrac%7BN%7D%7BC%7D%3D1.85%2A10%5E%7B12%7D%5Cfrac%7BN%7D%7BC%7D)