<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
The RDS-220 <span>hydrogen bomb, soviet </span>
Answer:
The radius of the gold nucleus is 7.1x10⁻¹⁵m
Explanation:
The nearest distance is:
(eq. 1)
Where
z = atomic number of gold = 79
e = electron charge = 1.6x10⁻¹⁹C
k = electrostatic constant = 9x10⁹Nm²C²
energy of the particle = 32 MeV = 5.12x10⁻¹²J
At the potential energy is zero, all the energy will be kinetic energy:
Where
m = 4 mp = mass of proton
Replacing in equation 1
Answer:
0.96 m
Explanation:
First, convert km/h to m/s.
162.3 km/h × (1000 m/km) × (1 hr / 3600 s) = 45.08 m/s
Now find the time it takes to move 20 m horizontally.
Δx = v₀ t + ½ at²
20 m = (45.08 m/s) t + ½ (0 m/s²) t²
t = 0.4436 s
Finally, find how far the ball falls in that time.
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.4436 s) + ½ (-9.8 m/s²) (0.4436 s)²
Δy = -0.96 m
The ball will have fallen 0.96 meters.
Answer:
659.01W
Explanation:
The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be
Wc = mass of the cab × acceleration due to gravity in m/s²
Wc = 1250 × 9.81 = 12262.5 N
but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N
Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N
the motor of the elevator will have to provide this in form of work
work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m
power provided by the motor of the elevator = workdone by the motor / time in seconds
Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W
