Answer:
11405Volt
Explanation:
To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.
The equation that allows the calculation of this voltage is given by,
Where
B = Magnetic field
A= Area
N = Number of loops
= Angular velocity
Our values previously given are:
We need convert the angular velocity to international system, then
Applying the equation for emf, we replace the values and we will obtain the value.
Answer:
Solvent
Explanation:
Answer:
Radient to ElEcTrIcAAl
Explanation:
The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.
Answer:
Force, F = −229.72 N
Explanation:
Given that,
First charge particle,
Second charged particle,
Distance between charges, d = 0.0359 m
The electric force between the two charged particles is given by :
F = −229.72 N
So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.
Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field. The work would it take to push two protons will be 7.7×10⁻¹⁴.
Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electricfield.
The given data in the problem is;
q is the charge= 1.6 ×10⁻¹⁹ C
V is the electric potential
r₁ is the first separation distance= 2.00×10−10 m
r₂ is the second separation distance= 3.00×10−15 m
The electric potential generated by the proton at rest at the two points, using the formula:
Firstly the electric potential at loction 1
The electric potential at loction 2
The product of difference of electric potential and charge is defined as the workdone.
Hence the work would it take to push two protons will be 7.7×10⁻¹⁴.
To learn more about electric potential work refer to the link.
We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?
The work done is equal to the electric potential energy gained by the proton:
where is the charge of the proton and is the potential difference between the final position and the initial position of the proton. To calculate this , we must calculate the electric potential generated by the proton at rest at the two points, using the formula:
where is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find
Therefore, the work done is
Answer:
Recoil velocity of cannon = 2.92 m/s
Explanation:
By law of conservation of momentum, we have momentum of cannon = momentum of cannonball.
Mass of cannon = 1200 kg
Mass of cannon ball = 100 kg
Velocity of cannon ball = 35 m/s
We have, Momentum of cannon = momentum of cannon ball
1200 x v = 100 x 35
v =3500/1200 = 2.92 m/s
Recoil velocity of cannon = 2.92 m/s
The recoil speed of the cannon is 2.92 m/s.
To find the recoil speed of the cannon, we can use the conservation of momentum. The initial momentum of the cannon and cannonball system is zero since the cannon is at rest before firing. The final momentum is the sum of the momenta of the cannon and cannonball after firing. Using the equation:
Initial momentum = Final momentum
(mass of cannon) x (recoil speed of cannon) = (mass of cannonball) x (velocity of cannonball)
Plugging in the given values:
(1200 kg) x (recoil speed of cannon) = (100 kg) x (35 m/s)
Solving for the recoil speed of the cannon:
recoil speed of cannon = (100 kg x 35 m/s) / 1200 kg = 2.92 m/s
#SPJ3
the right, and the initial velocity of the ball B is 6 meters per second to the left. The
final velocity of ball A is 9 meters per second to the left, while the final velocity of
ball B is 6 meters per second to the right.
1. Explain what happens to each ball after the collision. Why do you think this
occurs? Which of Newton’s laws does this represent?
Answer:
Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s and the total momentum after the collision is 1.5 kg • m/s. The momentum before and after the collision is the same.
Explanation: