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sergij07 [2.7K]
3 years ago
10

What is the electric current if a resistance of 100 Q and voltage of 12.0 V?

Physics
2 answers:
mamaluj [8]3 years ago
7 0

Answer:

letter A

Explanation:

yan po tamang sagot

sergeinik [125]3 years ago
5 0
We use the formula v=ir where I is current, v is voltage, and r is resistance, we get that r=12/100 which is answer choice A
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A cylindrical tungsten filament 16.0 cm long with a diameter of 1.00 mm is to be used in a machine for which the temperature wil
KonstantinChe [14]

Answer:

Explanation:

Resistance of the tungsten wire

R = resistivity x length / cross sectional area

= \frac{5.25\times10^{-8}\times16\times10^{-2}}{3.14\times(.5\times10^{-3})^2}

= 107 x 10⁻⁴ ohm

Resistance at 120 degree can be obtained from the following formula

R_t = R_0( 1 + \alpha\times t )

R_t = 107\times10^{-4}( 1 + .0045\times 100)

= 155.15 x 10⁻⁴ ohm

= 160 x 10⁻⁴ ohm ( rounding off to two syg fig )

current = 12.5

potential diff = 12.5 x 155.15 x 10⁻⁴ V

=  0 .1939 V

= .19 V

required electric field = potential diff / length of wire

= .1939 / 16 x 10⁻²

= 1.2 N / C

4 0
3 years ago
Read 2 more answers
Which choice would reveal a chemical property of water?
r-ruslan [8.4K]

Answer:

D

Explanation:

Testing The PH Means Testing The acidity level of the water of which would be a chemical reaction

8 0
3 years ago
A ball undergoes uniform acceleration of 2.50 m/s2 (beginning at rest). How far does the ball travel during that 4s time interva
-Dominant- [34]

Answer:

The ball travels a distance of 20 m in the time interval of 4 s

Explanation:

Using s = ut + 1/2at² where s = distance travelled by the ball, u = initial velocity of ball = 0 m/s (since it starts from rest), a = acceleration of the ball = 2.50 m/s² and t = time = 4 s.

Substituting the variables into the equation, we have

s = ut + 1/2at²

s = 0 × 4 s + 1/2 × 2.50 m/s² × (4 s)²

s = 0 + 1/2 × 2.50 m/s² × 16 s²

s = 1/2 × 40 m

s = 20 m

So, the ball travels a distance of 20 m in the time interval of 4 s.

4 0
4 years ago
If the net force acting on a stationary object is zero, then the object will A. remain at rest. B. accelerate in the direction o
ra1l [238]

The object will remain at rest

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

Here net force is zero

An object with a net force of zero acting on it will remain at rest, if initially at rest, or it will maintain a constant velocity.

F=MA

Force F=0

Than Acceleration A=0

With zero Acceleration stationary object  will remain at rest.

Hence Remain at rest is the correct answer

Learn more about Newton's law of motion here

brainly.com/question/25545050

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6 0
2 years ago
PLZ HELP ME
bonufazy [111]

Answer:

1. The stone will strike the ground 49.46 m from the base of the cliff

2. A) Approximately 0.542 seconds

B)  Approximately 3.69 m/s

3. A) The time the ball spends in the air is approximately 4.0775 s

B) The horizontal range is approximately  141.25 m.

Explanation:

1. The time it takes the stone to land is given by the equation, t = √(h/(1/2 × g)

∴ t = √(30/(1/2 × 9.81)) ≈ 2.473 seconds

The horizontal distance covered by the stone in that time = 20 × 2.473 ≈ 49.46 m

The stone will strike the ground 49.46 m from the base of the cliff

2. A) The time the ball spends in the air = t = √(h/(1/2 × g)

∴ t = √(1.44/(1/2 × 9.81)) ≈ 0.542 seconds

B) The initial horizontal velocity, u = Horizontal distance/(Time) = 2/0.542 ≈ 3.69 m/s

The initial horizontal velocity ≈ 3.69 m/s

3. A) The time the ball spends in the air is given by the following equation;

t = 2 × u × sin(θ)/g = 2 × 40 × sin(30)/9.81 ≈ 4.0775 s

t ≈ 4.0775 s

B) The horizontal range, R, of the  ball is given by the equation for the range of a projectile as follows;

Range, R = \dfrac{u^2 \times sin (2 \cdot \theta) }{g}

Substituting the known values, gives;

Range, R = \dfrac{40^2 \times sin (2 \times 30^{\circ}) }{9.81} \approx 141.25 \ m

The horizontal range ≈ 141.25 m.

4 0
3 years ago
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