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3 years ago

15

A 110 V power line is protected by a 15 A fuse. What is the maximum number of 400 W lamps that can be simultaneously operated in

parallel on this line without "blowing" the fuse because of an excess of current

1 answer:

Nitella [24]3 years ago

5 0

Answer:

Total number of lamps will be 4

Explanation:

We have given power of the lamp W = 400 watt

Potential difference across the lamp V=110 volt

We know that power is equal to $P=VI$

So $400=110\times I$

$I=3.636A$

Total current is given 15 A

As it is given that lamps are connected in parallel so total current is the sum of current through each lamp

So number of lamp will be $n=\frac{15}{3.636}=4.125$

As the lamp can not be in negative

So total number of lamps will be 4

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If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral

leonid [27]

Answer:

$\theta = tan^{-1}\mu$

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

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now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

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so we have

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3 years ago

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

tatuchka [14]

<u>Answer:</u> The final temperature of the solution is $80.14^oC$

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

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3 years ago

The second-order rate constants for the reaction of oxygen atoms ·with aromatic hydrocarbons have been measured (R. Atkinson and

Blizzard [7]

Answer: Frequency factor A = 8 × 10⁹

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Explanation: to begin, let us first define the parameters given;

K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹

K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹

K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹

also T₁ = 300.3 K

T₂ = 341.2 K

T₃ = 392.2 K

we know that;

㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]

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T₁, T₂ = Temperature

therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]

this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol

∴ Ea = 15.5 KJ/ Mol

also given that K = A e⁻∧Ea/RT

here A = frequency factor

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3 years ago

Which of these charges is experiencing the electric field with the largest magnitude? A 2C charge acted on by a 4 N electric for

Pavlova-9 [17]

Answer:

The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.

Explanation:

The formula for electric field is given as:

E = F/q

where,

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F = Electric Force

q = Charge Experiencing Force

Now, we apply this formula to all the cases given in question.

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F = 4 N

q = 2 C

Therefore,

E = 4 N/2 C = 2 N/C

B) <u>A 3 C charge acted on by a 5 N electric force</u>

F = 5 N

q = 3 C

Therefore,

E = 5 N/3 C = 1.67 N/C

C) <u>A 4 C charge acted on by a 6 N electric force</u>

F = 6 N

q = 4 C

Therefore,

E = 6 N/4 C = 1.5 N/C

D) <u>A 2 C charge acted on by a 6 N electric force</u>

F = 6 N

q = 2 C

Therefore,

E = 6 N/2 C = 3 N/C

E) <u>A 3 C charge acted on by a 3 N electric force</u>

F = 3 N

q = 3 C

Therefore,

E = 3 N/3 C = 1 N/C

F) <u>A 4 C charge acted on by a 2 N electric force</u>

F = 2 N

q = 4 C

Therefore,

E = 2 N/4 C = 0.5 N/C

The highest field is 3 N, which is found in part D.

<u>A 2 C charge acted on by a 6 N electric force</u>

3 0

3 years ago