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3 years ago

The formula for Ohm’s law can be written: V equals a. l x R b. l/R c. R/l

Physics

1 answer:

stellarik [79]3 years ago

7 0

The formula for Ohm’s law can be written: V equals l x R

Answer: Option A

<u>Explanation:</u>

Ohm's law defines that an electric current flowing through the conductor between two ends is directly proportionate to the voltage at these two points. The introduction of a constant proportionality, resistance, gives a simple mathematical equation describing this relationship. Particularly, Ohm's law also mentions that R is constant in this respect i.e. independent on current.

$\text {voltage }(V)=\text { current }(I) \times \text { Resistance }(R)$

Where,

I is the current conduction in amperes

V is the voltage calculated by the conductor in volts

R is the conductor’s resistance in ohms.

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In the 1660s, who discovered that color is a direct function of light by passing sunlight through a prism and observing the band

Mnenie [13.5K]

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5 0

3 years ago

6. several light bulbs are connected in series across a 115 v source

Natalka [10]

Explanation:

a) I = V / R

1.70 = 115 / R

R = 115 / 1.70

R = 67.647

R = 67.65 ohms

Therefore, equivalent resistance is 67.65 ohms

b) Equivalent resistance of circuit from above sum is 67.65 ohms

Given resistance of each bulb is 1.50 ohms

Number bulbs = Equivalent resistance / Resistance of each bulb

= 67.65 / 1.50

= 45

8 0

2 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$