Answer:
0.247 μC
Explanation:
As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:
The electric force is given by the expression:
In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):
And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.
O 0.247 μC
To solve this problem , we will use the length contraction equation as follows:
L = L0((1 - v^2/c^2))^0.5
where:
L is the desired length
L0 is the length measured at initial frame = 1 m
v is the velocity
c is the speed of light
Substituting by these givens in the above equation, we can calculate L as follows:
L = 1((1 - (0.8 c)^2/c2))^0.5 = 0.6 meters
Your hypothesis can be that the group with calculators would finish faster than the group without calculators.
The initial velocity of the 3250 Kg mass is 2.1 m/s. The distance covered by the larger mass in 5s is 4.7 cm.
In this problem, we have to apply the law of conservation of linear momentum. Note that;
Momentum before collision = Momentum after collision
m1u1 + m2u2 = m1v1 + m2v2
(2150 × 10) + (3250u1) = (2150 + 3250)5.22
21500 + 3250u1 = 5400 × 5.22
3250u1 = 28188 - 21500
u1 = 28188 - 21500/3250
u1 = 2.1 m/s
2) Again from the principle of conservation of linear momentum;
(0.40 × 3.5) + (0.60 × 0) = (0.40 × 0.70) + (0.60 × v2)
1.4 = 0.28 + 0.60v2
1.4 - 0.28 = 0.60v2
v2 = 1.87 cm/s
Using;
s = 1/2 ( u + v)t
s = 1/2(0 + 1.87) × 5
s = 4.7 cm
Learn more: brainly.com/question/8898885
The negative charges and positive charges in the wire are pushed in the opposite directions by the magnetic field