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Tems11 [23]
3 years ago
12

What voltage battery would you need to send 2.5A of current through a light bulb of resistance 3.6 ohm

Physics
1 answer:
nadezda [96]3 years ago
8 0
<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>

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YO CAN ANYONE DO THE BLANK COLUMN AND THE QUESTION PART RQ PLS!!
VikaD [51]

Answer:

stryo:  1

wood: 1

ice: 1

brick: 2

aluminum: 2.7

Explanation:

d= mass/ total volume

(fyi: for aluminum, they did the subtraction wrong to find the total volume. it is actually 5 or 5.00)

6 0
3 years ago
The x-component of vector R is Rx = −28.2 units and its y-component is Ry = 19.6 units. What are its magnitude and direction? Gi
ikadub [295]

Answer:

Explanation:

Rx = -28.2 units

Ry = 19.6 units

magnitude of R = √  [( - 28.2 )² + ( 19.6 ) ]

= √ ( 795.24 + 384.16 )

= 34.34 units

If θ  be the angle measured counterclockwise from the +x-direction

Tanθ = 19.6 / - 28.2 = -0.695

θ = 180 - 34.8

= 145.2° .

7 0
3 years ago
If a fisherman applies a horizontal force with magnitude 47.0 NN to the box and produces an acceleration of magnitude 3.20 m/s2m
musickatia [10]

Answer:

The correct solution is "14.6875 kg".

Explanation:

Given values:

Force,

F = 47.0 N

Acceleration,

a = 3.20 m/s²

Now,

⇒ Force=Mass\times Acceleration

or,

⇒       F=ma

⇒    47.0=m\times 3.20

⇒       m=\frac{47.0}{3.20}

⇒           =14.6875 \ kg

4 0
2 years ago
If you are stopped, then get on a bike and accelerate 2 m/s/s south, what us your final velocity after 3 seconds? Please Explain
olga_2 [115]
Vf = Final velocity.
Vi = initial velocity
a = acceleration. 
t = time

Vf = Vi + at

Vf = 0 + (2 m/s^2)(3s)

Vf = 6 m/s south
4 0
3 years ago
A track is traveling a. a speed of 25.0 m/s along a level road. A crate is resting on the bed of the truck, and the coefficient
Mice21 [21]

To solve this problem it is necessary to apply the concepts related to

conservation of energy, for this case manifested through work and kinetic energy.

W = \Delta KE

W = F*d

Where,

F= Force (Frictional at this case F_r = \mu N)

d= Distance

\Delta KE = \frac{1}{2} mv^2

Where,

m = mass

v = velocity

Equation both terms,

F*d = \frac{1}{2}mv^2

\mu mg *d = \frac{1}{2}mv^2

\mu g * d = \frac{1}{2}v^2

d = \frac{1}{2} \frac{v^2}{\mu g}

Replacing with our values we have that

d = \frac{1}{2} \frac{25^2}{0.65*9.8}

d = 49.05m

Therefore the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck is 49.05m

6 0
3 years ago
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