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Zepler [3.9K]
3 years ago
8

Explain the difference between a fuse and a circuit breaker if you were designing a circuit for a reading lamp would you include

a fuse a circuit breaker or neither
Physics
1 answer:
sesenic [268]3 years ago
7 0
The answer is Fuse.  Circuit breakers and fuses are generally known as "Overload Protection Devices.  The difference between a fuse and a circuit breaker is that a fuse is a one time circuit element while a circuit breaker is a multiple use device. If you were designing a circuit for a reading lamp, a fuse would do.  Since almost all household installed a circuit breaker for outlets, etc,
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Relate the structure of water to its ability to act as a solvent
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Explanation:

Water (H2O) as a polar covalent molecule has its arrangement of oxygen and hydrogen atoms where one end (hydrogen) has a partially positive charge while the other side (oxygen) had a partially negative charge.

It is also capable of forming hydrogen bonds with polar molecules. Each water molecule can form two hydrogen bonds involving their hydrogen atoms and two further hydrogen bonds using the hydrogen atoms attached to neighboring water molecules.

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It is cold and dry outside. You go down the slide and experience a small electric shock. What charge must you and slide be in or
ruslelena [56]

Answer:

c

Explanation:

the charge wants to ultimately balance out so either you or the slide gave the other electrons to balance the charges

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What do crystal of table salt look like?
Pavel [41]

They look like tiny cubes. They are in a cubic form.


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How much Hydrogen (H) atoms are in this 5NH4Cl?
Jobisdone [24]

Answer:

5

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8 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
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