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GrogVix [38]
3 years ago
8

PLEASE HELP QUICK ASAP HURRY This is Earth Science

Physics
1 answer:
SCORPION-xisa [38]3 years ago
4 0
Sun this is the answer
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What voltage battery would you need to send 2.5A of current through a light bulb of resistance 3.6 ohm
nadezda [96]
<h3><u>Given</u> :</h3>

Current flow light bulb = 2.5\sf{A}

Resistance of light bulb = 3.6Ω

<h3><u>To Find </u>:</h3>

We have to find voltage of battery.

<h3><u>Solution</u> :</h3>

➠ As per ohm's law, current flow through a conductor is directly proportional to the applied potential difference.

➝ V ∝ I

➝ <u>V = I × R</u>

Where, R is the resistance of conductor.

⇒ V = I × R

⇒ V = 2.5 × 3.6

⇒ <u>V = 9 volt</u>

8 0
3 years ago
The acceleration of an object is constant when its velocity is :
emmasim [6.3K]

Answer:

B. changing by a constant amount each second

Explanation:

thats my answer

5 0
3 years ago
Read 2 more answers
A large pendulum with a 200-lb gold plated bob 12 inches in diameter is on display in the lobby of the united nations building.
BaLLatris [955]
The period of one full swing depends on the length of the pendulum and on gravity. The period of each full swing would be longer on the moon, with less gravity.

The rotation of the plane of the swings doesn't depend on the length of the string OR on gravity. It only depends on the latitude of the place where the pendulum hangs, and the rotation period of the body it's located on.

On Earth, it's (24 hours)/(sine of latitude).

On the moon, it would be (27.32 days)/(sine of latitude).
8 0
3 years ago
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When the distance between two stars decreases by one-third, the force between them
pashok25 [27]

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

r'=\frac{2}{3}r

So the new force will be

F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F

So, the force will be 2.25 times the previous value.

4 0
4 years ago
A 150kg motorcycle starts from rest and accelerates at a constant rate along with a distance of 350m. The applied force is 250N
mart [117]

Answer:

205N

Explanation:

The net force (F) is the difference between the applied force(F_{A}) and the kinetic frictional force(F_{R}). i.e

F = F_{A} - F_{R}    -----------------(i)

Note that;

F_{R} = μmg

Where;

μ = coefficient of kinetic friction

m = mass of the body

g = acceleration due to gravity = 10m/s²

Equation (i) then becomes;

F = F_{A} - μmg        -------------------(ii)

<em>Given from question;</em>

m = mass of motorcycle = 150kg

μ = 0.03

F_{A} = 250N

Substitute these values into equation (ii) as follows;

F = 250 - (0.03 x 150 x 10)

F = 250 - (45)

F = 205N

Therefore, the net force applied to the motorcycle is 205N

3 0
3 years ago
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