1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Montano1993 [528]
2 years ago
6

How can you show that an electric current has a magnet affect​

Physics
1 answer:
lubasha [3.4K]2 years ago
5 0

Answer:

It is observed that when a compass is brought near a current carrying conductor the needle of compass gets deflected because of flow of electricity. This shows that electric current produces a magnetic effect.

You might be interested in
Explain why theories may be changed or replaced overnight time.
vlabodo [156]
They may be changed because they may find evidence of some thing that will change their perspective on things.
5 0
3 years ago
A baseball pitcher loosens up his pitching arm. He tosses a 0.20-kg ball using only the rotation of his forearm, 0.28 m in lengt
aev [14]

Answer:

Moment of Inertia, I = 0.016 kgm²

Explanation:

Mass of the ball, m = 0.20 kg

Length of the pitcher's arm, l = 0.28

Radius of the circular arc, r = 0.28 m

Moment of Inertia is given by the formula:

I = mr²

I = 0.20 * 0.28²

I = 0.20 * 0.0784

I = 0.01568

I = 0.016 kgm²

7 0
3 years ago
1. Determine the magnitude of two equal but opposite charges if they attract one another with a force of 0.7N when at distance o
adell [148]

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

        F =k \frac{q_1q_2}{r^2}

In this case they indicate that the load is of equal magnitude

       q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

       F = k \ \frac{q^2}{r^2}

       q = \sqrt{\frac{F \ r^2}{k} }

we calculate

        q = \sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9}  }

        q = \sqrt{7 \ 10^{-12} }Ra 7 10-12

        q = 2.65 10⁻⁶ C

7 0
3 years ago
A particle with a mass of 6.64 × 10–27 kg and a charge of +3.20 × 10–19 C is accelerated from rest through a potential differenc
blondinia [14]

Answer:

Explanation:

Given that,

Mass m = 6.64×10^-27kg

Charge q = 3.2×10^-19C

Potential difference V =2.45×10^6V

Magnetic field B =1.6T

The force in a magnetic field is given as Force = q•(V×B)

Since V and B are perpendicular i.e 90°

Force =q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I.e K•E=U

K•E =½mV²

K•E =½ ×6.64×10^-27 V²

K•E = 3.32×10^-27 V²

U = q•V

U = 3.2×10^-19 × 2.45×10^6

U =7.84×10^-13

Then, K•E = U

3.32×10^-27V² = 7.84×10^-13

V² = 7.84×10^-13 / 3.32×10^-27

V² = 2.36×10^14

V=√2.36×10^14

V = 1.537×10^7 m/s

So, applying this to force in magnetic field

F=q•V•B

F= 3.2×10^-19 × 1.537×10^7 ×1.6

F = 7.87×10^-12 N

6 0
3 years ago
Read 2 more answers
Can someone help me out?
MArishka [77]

Answer:

B). Rarefaction :) _____

8 0
3 years ago
Read 2 more answers
Other questions:
  • Select the best terms from the drop-down menus that fit the nuclear medicine descriptions.
    9·2 answers
  • Hey! Can someone help with this question? Thx :)
    10·1 answer
  • In the graph, which two regions show the particle undergoing zero acceleration and negative acceleration respectively?
    11·1 answer
  • A student throws a rock horizontally off a 5.0 m tall building. The rock's initial speed is 6.0 m/s. How long will it take the r
    8·2 answers
  • A neutral atom can be negatively charged by
    15·1 answer
  • What is a symptom?
    7·1 answer
  • Janet jumps horizontally off a high diving platform with a velocity of 2.63 m/s and lands in the water 1.9 s later. How high is
    8·1 answer
  • 1:A pattern or grouping of stars that people imagine representing a figure, an animal or an object.
    14·1 answer
  • Please help will mark brainliest!
    15·2 answers
  • I need help with 03.01 Playing With Others – President's Challenge Goals
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!