Answer: q2 = -0.05286
Explanation:
Given that
Charge q1 = - 0.00325C
Electric force F = 48900N
The electric field strength experienced by the charge will be force per unit charge. That is
E = F/q
Substitute F and q into the formula
E = 48900/0.00325
E = 15046153.85 N/C
The value of the repelled second charge will be achieved by using the formula
E = kq/d^2
Where the value of constant
k = 8.99×10^9Nm^2/C^2
d = 5.62m
Substitutes E, d and k into the formula
15046153.85 = 8.99×10^9q/5.62^2
15046153.85 = 284634186.5q
Make q the subject of formula
q2 = 15046153.85/ 28463416.5
q2 = 0.05286
Since they repelled each other, q2 will be negative. Therefore,
q2 = -0.05286
Answer:
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Explanation:
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On a similar problem wherein instead of 480 g, a 650 gram of bar is used:
Angular momentum L = Iω, where
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>
<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>
<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>
<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>
<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
The work done by the machine is equal to the product between the force applied and the distance over which the force is applieds, so in this case:

And the power of the machine is equal to the ratio between the work done by the machine and the time taken:
<em>Resultant angle; θ = 25.59° </em>
This question is dealing with bearings and distance.
We are told that from point A, the camel walks 20 km at 15° in the south of east direction.
Thus, d_s,e = 20 km
Resolving along the horizontal east direction gives; d_e = 20 cos 15
d_e = 19.32 km
Also, resolving along the vertical south direction gives; d_s = 20 sin 15
d_s = 5.18 km
Net vertical distance; d_vert = 8km - 5.18km = 2.72 km
Net horizontal distance; d_hor = 25km - 19.32 km = 5.68 km
Now, the resultant angle is given by;
tan θ = d_vert/d_hor
tan θ = 2.72/5.68
tan θ = 0.4789
θ = tan^(-1) 0.4789
θ = 25.59°
Read more at; brainly.com/question/22518031