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olganol [36]
3 years ago
12

If you set up an experiment with two different independent variables, then the results would be​

Physics
1 answer:
valina [46]3 years ago
3 0

Answer: the same

Explanation:

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What would be the magnitude of the electric field 0.75 m from a 0.63 C master charge and what would be the force on a 0.50 C tes
bagirrra123 [75]

The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

<h3>Electric field on the master charge</h3>

E = kq/r²

where;

  • q is magnitude of master charge
  • r is distance of separation
  • k is Coulomb's constant

E = (9 x 10⁹ x 0.63)/(0.75²)

E = 1.008 x 10¹⁰ N/C

<h3>Force on the test charge</h3>

F = Eq

where;

  • E is electric field
  • q is the test charge

F = (1.008 x 10¹⁰) x (0.5)

F = 5.04 x 10⁹ N

Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

8 0
1 year ago
(Astronomy)
olya-2409 [2.1K]

Answer:c ITs C ^w^

Explanation:

4 0
3 years ago
Help!!! Line B touches the circle at a single point. Line A extends through the center of the circle.
Ulleksa [173]

Answer:

If I understand correctly. Line B is parallel to the circle. Also, the angle is less than 90.

  1. The size of the circle determines.
  2. The diameter should not be fixed either.
3 0
3 years ago
A person's speed around the Earth is faster at the poles than it is at the equator.
VashaNatasha [74]

Answer:

no

Explanation:

it is faster at the equator

6 0
2 years ago
A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc
nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

m_sv_s + m_rv_r = 0

where m_s = 72.8, m_r = 265 are the mass of the swimmer and raft, respectively. v_s = 5.21 m/s, v_r are the velocities of the swimmer and the raft after the run, respectively. We can solve for v_r

265v_r + 72.8*5.21 = 0

v_b = -72.8*5.21/265 = -1.43 m/s

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

7 0
3 years ago
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