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2 years ago

How must a fuse be connected in a circuit to prevent current from flowing when the circuit becomes ""overloaded""?

Physics

1 answer:

Lady bird [3.3K]2 years ago

3 0

Answer:

The fuse must be connected between the device and the power intake source.

Explanation:

A fuse is a protective component of electrical appliances that is designed to be sensitive to a particular range of electric current

The fuse is made of a thing metal strip with a known melting point. Once current abive its carrying capacity flows through it, large heat is generated in the metal strip which melts it and causes the metal strip to cut int two protecting the device from the power spike.

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I need help, ASAP i’m failing and i have no clue what’s going on in my AP physics class at all.

garri49 [273]

What’s the question or problem ?

6 0

3 years ago

As part

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$

Part b)

$a = 11.8 m/s^2$

Explanation:

Part a)

For force conditions of two blocks we will have

$m_1g - T = m_1 a$

$T - m_2g = m_2 a$

now from above equations we have

$(m_1 - m_2) g = (m_1 + m_2) a$

$a = \frac{m_1 - m_2}{m_1 + m_2} g$

now we know that

$m_1 = \frac{908}{9.8} = 92.65 kg$

$m_2 = \frac{412}{9.8} = 42 kg$

now from above equation we have

$a = \frac{92.65 - 42}{92.65 + 42}(9.8)$

$a = 3.68 m/s^2$

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

$F - mg = ma$

$908 - 412 = 42 a$

$a = 11.8 m/s^2$

8 0

3 years ago

Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

Sedbober [7]

Answer:

(A) ratio of electric force to weight will be $23.469\times 10^{-10}$

(b) Electric field will be $E=4.26\times 10^{10}N/C$

Explanation:

We have given mass of bee = 100 mg = $m=100\times 10^{-3}=0.1kg$

Charge on bee $q=23pC=23\times 10^{-12}C$

Electric field E = 100 N/C

Weight of the bee $W=mg=0.1\times 9.8=0.98N$

Electric force on the bee $F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N$

So the ratio of electric force on the bee and weight is $=\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}$

(B) To hold the bee in air electric force must be equal to weight of bee

So $mg=qE$

$0.1\times 9.8=23\times 10^{-12}E$

$E=4.26\times 10^{10}N/C$

4 0

3 years ago

A man of mass 70 kg climbs stairs of vertical height 2.5m. Calculate the work done against the force of gravity. (Take g = 9.81

Alona [7]

Answer:

1716.75 J

Explanation:

Step 1: First check what we are provided with. As per given question we have:

mass (m) = 70 kg, height (h) = 2.5 m and acceleration due to gravity (g) = 9.81 m/s².

Step 2: Check what we are asked to find out.

Work done = Change in Potential energy

The stuff required to solve this question is potential energy. Using the formula: P = mgh. Where P is Potential energy, m is mass, g is acceleration due to gravity and h is height.

Step 3: Substitute the known values in the above formula.

→ P = 70 × 2.5 × 9.81

→ P = 1716.75 J

Hence, the work done against the force of gravity is 1716.75 J.

3 0

1 year ago

A gas in a cylinder with a moveable piston has an initial volume of 89.2 ml . if we heat the gas from 35 âc to 153 âc, what is i

Korvikt [17]

Use Charles' Law:
v1/T1 = v2/T2
88.2/(273+35) = v2/(273+155)
where v2 is the new volume.

I hope this helps!!!

4 0

2 years ago