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Svetradugi [14.3K]

2 years ago

8

In a house, 3 bulbs of 60 watt each are lighted for 3 hours daily, 4 fans of 100 watt each are used for 8 hours daily and an ele

ctric heater of 2 kwh used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of Rs 4/kwh.

1 answer:

Virty [35]2 years ago

5 0

Answer:

Total Energy consumed = 146.94 kwh

Total cost = rs 587.76

Explanation:

Given:

3 bulbs each 60 watt [for 3 hours daily]

4 fans each 100 watt [for 8 hours daily

1 electric heater of 2 kWh (2,000 watt) [for half hour daily ]

Rate = rs 4/Kwh

Find:

Total Energy consumed

Total cost

Computation:

Energy used = Power × time

3 bulbs each 60 watt [for 3 hours daily] = 3 x 60 x 3 x 31 = 16,740 watt = 16.74 kw

4 fans each 100 watt [for 8 hours daily = 4 x 100 x 8 x 31 = 99,200 watt = 99.2 kw

1 electric heater of 2 kWh [for half hour daily ] = 1 x 2000 x (1/2) x 31 = 31,000 watt = 31 kw

Total Energy consumed = 16.74 + 99.2 + 31

Total Energy consumed = 146.94 kwh

Total cost = 146.94 x 4

Total cost = rs 587.76

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Answer:

Conductors allow electric charges to move freely

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Answer:

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Explanation:

The computation of the displacement of the object is shown below:

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Hence, the displacement of the object is 5 units

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2 years ago

A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

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Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

2 years ago

A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is $m_r = 2 \ kg$

The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

$I_c = -6 \ kg\cdot \ m/s$

=> $| I_c| = 6 \ kg\cdot \ m/s$

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

8 0

3 years ago

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final

Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B = -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

$x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s$

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

5 0

2 years ago