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When three identical bulbs of 60W,200V rating are connected in series at a 200V supply the power drawn by them will be?

Law Incorporation [45]

Answer:

P = 180 [w]

Explanation:

To solve this problem we must use ohm's law, which is defined by the following formula.

V = I*R & P = V*I

where:

V = voltage = 200[volts]

I = current [amp]

R = resistance [ohm]

P = power [watts]

Since the bulbs are connected in series, the powers should be summed

P = 60 + 60 + 60

P = 180 [watts]

Now we can calculate the current

I = 180/200

I = 0.9[amp]

Attached is an image where we see the three bulbs connected in series, in the circuit we see that the current is the same for all the elements connected to the circuit.

And the power is defined by P = V*I

we know that the voltage is equal to 200[V], therefore

P = 200*0.9

P = 180 [w]

8 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

Describe The energy conversions that occur when coal is formed, and then burned to produce thermal energy.

svlad2 [7]

Answer:

When energy contained in coal is turned into heat, and then into electrical energy. As boiling water heated by the burning coal is cooled, steam forges from these cone-shaped cooling towers.

8 0

3 years ago

Which of the following is a result of a change in pressure?

ioda

B. Exfoliation. Hope I helped you out bro.

7 0

3 years ago

A trampoline spring has a force constant k = 800 N/m and is stretched exactly 17.5cm. What is the energy required to do this?

Artist 52 [7]

Answer:

the energy required for the extension is 12.25 J

Explanation:

Given;

force constant of trampoline spring, k = 800 N/m

extension of trampoline spring, x = 17.5 cm = 0.175 m

The energy required for the extension is calculated as;

E = ¹/₂kx²

E = 0.5 x 800 x 0.175²

E = 12.25 J

Therefore, the energy required for the extension is 12.25 J

6 0

3 years ago