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2 years ago

Suppose a family replaces ten 60-watt incandescent bulbs with ten 30-watt

Physics

1 answer:

mixer [17]2 years ago

5 0

Initially there were 10 bulbs of 60 Watt power

So total power of all bulbs = 60 * 10 = 600 W

now each bulb used for 4 hours daily

so total energy consumed daily

E = p*t

E = 600*4 = 2400Wh

E = 2.4kWH

now we have total power consumed in 1 year

E = 365 * 2.4 = 876kWh

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

P1 = 876*10 = 8760cents = 87.60

Now if all 60 Watt bulbs are replaced by 30 Watt bulbs

So total power of all bulbs = 30 * 10 = 300 W

now each bulb used for 4 hours daily

so total energy consumed daily

(The next steps are in the pic.)

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A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

3 years ago

How can a cyclist minimize friction as he or she rides? A. by increasing the weight of the bike B. by increasing the tread on hi

jek_recluse [69]

B I believe is the answer!

Hope this helps and have a great day!!!

6 0

2 years ago

Read 2 more answers

A bomb of mass 6kg initially at rest explodes into two fragments of masses 4kg and2kg respectively. If the greater mass moves wi

Katen [24]

Answer:

v = 10 [m/s]

Explanation:

The largest mass is that of 4 [kg], in this way the momentum can be calculated by means of the product of the mass by velocity.

$P=m*v\\$

where:

P = momentum [kg*m/s]

m = mass = 4 [kg]

v = velocity = 5 [m/s]

Now the momentum:

$P=4*5\\P=20[kg*m/s]$

This same momentum is equal for the other mass, in this way we can find the velocity.

$P=m*v\\20=2*v\\v=10[m/s]$

7 0

3 years ago

Consider an old-fashion bicycle with a small wheel of radius 0.17 m and a large wheel of radius 0.92 m. Suppose the rider starts

Gala2k [10]

Answer:

10259.6 m

Explanation:

We are given that

Radius of small wheel,r=0.17 m

Radius of large wheel,r'=0.92 m

Initial velocity,u=0

Time,t=2.7 minutes=162 s

1 min=60 s

Velocity,v=10m/s

Time,t'=13.7 minutes=822 s

Time,t''=4.1 minutes=246 s

$v=u+at$

Substitute the values

$10=0+162a=162a$

$a=\frac{10}{162}=0.0617m/s^2$

$s=ut+\frac{1}{2}at^2$

Substitute the values

$s=\frac{1}{2}(0.0617)(162)^2=809.6 m$

$s'=vt'=10\times 822=8220 m$

$a'=\frac{v}{t''}=\frac{10}{246}$

$s''=\frac{1}{2}a't''^2=\frac{1}{2}\times \frac{10}{246}(246)^2=1230 m$

Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m

6 0

3 years ago

Read 2 more answers

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

3 years ago