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garik1379 [7]
3 years ago
5

Fuses and circuit breakers are connected in ___?___ with the load and all other components in the circuit.

Physics
1 answer:
LuckyWell [14K]3 years ago
8 0

Answer:

series

Explanation:

the fuse and circuit breakers  are protective devices meant to protect all loads and components in a circuit, therefore current has to pass trough them before getting to the loads and components and hence they are connected in series. If current that comes trough the fuse is above its required rating, the fuse will cut and if its the breaker it will trip off, hence preventing flow of current and damage to other loads and circuitry.    

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The tail consists of gas and dust that can extend hundreds of millions of kilometers away from the coma. Most comets actually have 2 tails: a plasma tail made of ionizedgas, and a dust tail made of small solidparticles. Comet tails point away from the Sun.
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3 years ago
What is an important trait or skill a scientist must have
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Characteristics that lend themselves to carrying out the kind of work scientists do. These include strong observational skills, curiosity<span>, logic, </span>creativity<span>, skepticism, and objectivity.</span>
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4 years ago
A sprinter accelerates at 7.5 m/s from rest in 2.0 s, what distance did she go? (15 m)
a_sh-v [17]

Answer:

<em>The sprinter traveled a distance of 7.5 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the rate of change of the velocity of an object is constant.

The equation that rules the change of velocities is:

v_f=v_o+at\qquad\qquad [1]

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

Using the equation [1] we can solve for a:

\displaystyle a=\frac{v_f-v_o}{t}

The sprinter travels from rest (vo=0) to vf=7.5 m/s in t=2 s. Computing the acceleration:

\displaystyle a=\frac{7.5-0}{2}

a=3.75\ m/s^2

Now calculate the distance:

\displaystyle x=0*2+\frac{3.75*2^2}{2}

\displaystyle x=7.5\ m

The sprinter traveled a distance of 7.5 m

8 0
3 years ago
A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc
Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0
3 years ago
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Answer:

Explanation:

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