Explanation:
It is given that,
Diameter of loops, d = 48 cm = 0.48 m
Radius of the loop, r = 0.24 m
Current carried in the loop, I = 4.5 A
Distance between loops, x = 30 cm = 0.3 m
Speed of the proton, v = 2600 m/s
We know that the magnetic field at the midway of the coils is given by :
Let F is the magnetic force these loops exert on the proton just after it is fired. It is given by :
So, the magnetic force these loops exert on the proton is . Hence, this is the required solution.
Answer:
Yes . Frictions was not included for the calculation for acceleration .
Explanation:
This is because we're considering the weight of an object which isn't acceleration i:e frictionless.
If friction is included, the equation for acceleration becomes.
F= ma,(Newton second law of motion)
where F=frictional force , m =mass of the object and a=acceleration..
So therefore, a = F/m.
<span>3.78 m
Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes.
7.2 m/s / 9.81 m/s^2 = 0.77945 s
The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving
d = 1/2 A T^2
d = 1/2 9.81 m/s^2 (0.77945 s)^2
d = 4.905 m/s^2 0.607542 s^2
d = 2.979995 m
So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height.
d = 2.979995 m + 0.8 m = 3.779995 m
Rounding to 2 decimal places gives us 3.78 m</span>
Answer:
1.47×10¯⁸ N
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 1.5×10¯⁹ Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
The force of the snowflake can be obtained by using the following formula:
F = mg
Where:
F => is the force
m => is the mass
g => is the acceleration due to gravity
F = mg
F = 1.5×10¯⁹ × 9.8
F = 1.47×10¯⁸ N
Thus the force of the snowflake is 1.47×10¯⁸ N
The total circuit current at the resonant frequency is 0.61 amps
What is a LC Circuit?
- A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
- These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.
Q =15 = (wL)/R
wL = 30 ohms = Xl
R = 2 ohms
Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance
| Zs | = 30.07 <86.2° ohms
Xc = 1/(wC) = 30 ohms
The impedance of the LC circuit is found from:
Zp = (Zs)(-jXc)/( Zs -jXc)
Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°
I capacitor = 277/-j30 = j9.23 amps
I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps
I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps
Hence, the total circuit current at the resonant frequency is 0.61 amps
To learn more about LC Circuit from the given link
brainly.com/question/29383434
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