Explanation:
F = MA
200 = 100 * A
A = 200/100
A = 2m/sec^2
<h3><em>hope </em><em>it </em><em>helps </em><em>you </em></h3>
Answer:
the question is incomplete, the complete question is
"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"
2.6mA
Explanation:
we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.
using the formula be low,
![E=-\frac{d}{dt}(BACOS\alpha )\\](https://tex.z-dn.net/?f=E%3D-%5Cfrac%7Bd%7D%7Bdt%7D%28BACOS%5Calpha%20%29%5C%5C)
where B is the magnitude of the field and A is the area of the circular coil.
First, let determine the area using
where r is the radius of 5cm or 0.05m
![A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\](https://tex.z-dn.net/?f=A%3D%5Cpi%20%2A%280.05%29%5E%7B2%7D%5C%5C%20A%3D0.00785m%5E%7B2%7D%5C%5C)
since we no that the angle is at
we determine the magnitude of the magnetic filed
![B=0.5e^{-t} \\t=2s](https://tex.z-dn.net/?f=B%3D0.5e%5E%7B-t%7D%20%5C%5Ct%3D2s)
![E=-0.000532v\\](https://tex.z-dn.net/?f=%20E%3D-0.000532v%5C%5C)
the Magnitude of the voltage is 0.000532V
Next we determine the current using ohm's law
![V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A](https://tex.z-dn.net/?f=V%3DIR%5C%5CR%3D0.2%5C%5CI%3D%5Cfrac%7B0.000532%7D%7B0.2%7D%20%5C%5CI%3D0.0026A)
![I=2.6mA](https://tex.z-dn.net/?f=I%3D2.6mA)
<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>
When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a measure of the ability of a capacitor to store energy. </em>For any capacitor, the capacitance is a constant defined as:
![C=\frac{Q}{V_{ab}}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7BQ%7D%7BV_%7Bab%7D%7D)
To maintain
constant, if Q increases V also increases.
On the other hand, the potential energy
can be expressed as:
![U=\frac{Q^{2}}{2C}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7BQ%5E%7B2%7D%7D%7B2C%7D)
In conclusion, as Q increases the potential energy also increases.
Answer:
f1 = -3.50 m
Explanation:
For a nearsighted person an object at infinity must be made to appear to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.
∴ v = -3.5 m
Using mirror formula,
i/f1 = 1/v + 1/u
Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u = object distance = at infinity(∞) = 1/0
∴ 1/f1 = (1/-3.5) + 1/infinity
Note that, 1/infinity = 1/(1/0) = 0/1 =0.
∴ 1/f1 = 1/(-3.5) + 0
1/f1 = 1/(-3.5)
Solving the equation by finding the inverse of both side of the equation.
∴ f1 = -3.50 m
Therefore a converging lens of focal length f1 = -3.50 m
would be needed by the person to see an object at infinity clearly
Answer:
i guess 0.8 miles away
Explanation:
mathematically:-
speed is given in question
Time is given
and we have to find distance
simply by using speed formula (s) = d/t
we get answer