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velikii [3]
3 years ago
10

a 20 A fuse is connected in series with a circuit containing a 240 V source. What is the minimum resistance required to prevent

the fuse from melting
Physics
1 answer:
kipiarov [429]3 years ago
7 0

Answer:

12 ohms

Explanation:

you just divide 240 and 20

240 V/ 20 A = 12 ohms

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How are the sun, the moon, and Earth related during a solar eclipse?
german

B. The moon is located between the Sun and Earth

4 0
3 years ago
What is the amount of heat, in Sl units, necessary to melt 1 lb of ice?
likoan [24]

Answer:

Heat required to melt 1 lb of ice is 151.469 KJ

Explanation:

We have given mass of ice = 1 lb

We know that 1 lb = 0.4535 kg

Latent heat of fusion for ice =334 KJ/kg

Amount if heat for fusion of ice is given by

Q=mL, here m is mass of ice and L is latent heat of fusion

So heat Q=mL=0.4535\times 334=151.469kj

So heat required to melt 1 lb of ice is equal to 151.469 KJ

3 0
3 years ago
The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t
slamgirl [31]

Answer:

The charge is 2.75\times10^{-13}\ C

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length L=5.0\times5.0\times10^{-4}\ m

We need to calculate the linear charge density

Using formula of linear charge density

E=\dfrac{2k\lambda}{r}

\lambda=\dfrac{Er}{2k}

Put the value into the formula

\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}

\lambda=1.1\times10^{-10}\ C/m

We need to calculate the charge

Using formula of charge

Q=\lambda\timesL

Put the value into the formula

Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})

Q=2.75\times10^{-13}\ C

Hence, The charge is 2.75\times10^{-13}\ C

5 0
3 years ago
The gravitational force between two objects that
leonid [27]

Answer:

The answer to your question is    m₂ = 38.5 kg

Explanation:

Data

distance = d = 2.1 x 10⁻¹ m

Force = 3.2 x 10⁻⁶ N

m₁ = 55 kg

m₂ = ?

G = 6.67 x 10 ⁻¹¹ Nm²/kg²

Process

1.- To solve this problem use Newton's law of Universal Gravitation.

             F = G m₁m₂ / r²

-Solve for m₂

            m₂ = Fr² / Gm₁

2.- Substitution

            m₂ = (3.2 x 10⁻⁶)(2.1 x 10⁻¹)² / (6.67 x 10⁻¹¹)(55)

3.- Simplification

            m₂ = 1.411 x 10⁻⁷ / 3.669 x 10⁻⁹

4.- Result

            m₂ = 38.5 kg

5 0
3 years ago
In terms of saturation (unsaturated, saturated, super-saturated). How would you classify the following?
miskamm [114]

<u>Answer:</u>

<em>1. A NaCl solution with a concentration of 50g/100mL of water at 40°C:</em> The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.

<em>2. A sugar solution with a concentration of 200g/100mL of water at 40°C: </em>The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.

<em>3. A sugar solution with a concentration of 240g/100mL of water at 40°C:</em> The sugar solution with a given concentration is saturated at given temperature.

7 0
3 years ago
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