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amid [387]
2 years ago
12

what is the name of the area around a charged object where the object can exert a force on other charged objects?

Physics
1 answer:
shusha [124]2 years ago
4 0

Answer:

An electric field is a region around a charged object where the object's electric force is exerted on other charged objects. Electric fields get weaker the farther away they are from the charge. An electric field is invisible. You can use the field line to represent it.

Explanation:

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Why is an element considered a pure substance?
anastassius [24]

Answer:

An element is a pure substance and is made of only one type of atom.

Explanation:

It cannot be broken down into a simpler substance.

3 0
2 years ago
Question is down below
rosijanka [135]

The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s

<h3>What are the components of velocity?</h3>

We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.

Hence;

Vertical component = 40 m/s sin 15 degrees = 10.35 m/s

Horizontal component = 40 cos 15 degrees = 38.6 m/s

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7 0
1 year ago
G a magnetic field perpendicular to the plane of a wire loop is uniform in space but changes with time t in the region of the lo
Luden [163]

Answer:

e = Δφ / Δt     induced emf is proportional to enclosed flux

Also φ  = B * A      flux is proportional to area and enclosed field

If the induced emf e increases with time than the flux and hence the magnetic field is increasing with time  (replace B with G)

Since e = ΔG * A / Δt    if e is linear then G must also be linear and be proportional to the time

6 0
1 year ago
When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6
Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = +8.4\mu C

q_2 = +5.6 \mu C

force between two charges is given as

F = 0.66 N

now we have

F = \frac{kq_1q_2}{r^2}

0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}

r = 0.8 m

so it is separated by 80 cm distance

3 0
2 years ago
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Answer:

B. fitness goal I believe. please like this, I need points. ty!

Explanation:

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