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BartSMP [9]
4 years ago
7

A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current

running through the entire circuit?
Physics
1 answer:
larisa [96]4 years ago
6 0

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega

And now we can use Ohm's law to find the current in the circuit:

V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A

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How can artists suggest volume with light?
Dimas [21]

Answer:

The answer is they utilized light in a few spots to enlighten the subjects and to demonstrate their volume as masses.  

Explanation:

That is the reason every one of the particles that move at the speed of light (e.g. photons) have zero rest mass. As a molecule with mass methodologies the speed of light, its vitality increments and winds up unbounded at the speed of light, which is the motivation behind why it can never be quickened to achieve that speed. However, you can state that the photon has relativistic mass on the off chance that you truly need to. In current wording, the mass of a protest is its invariant mass, which is zero for a photon.

5 0
3 years ago
A 2kg box is pushed along a flat frictionless surface with an applied force of 53.91newton j+19.62 j were j is horizontal and j
tamaranim1 [39]

Since the box doesn't move vertically at all, no work is done by the vertical component of the force.

If 53.91 Newtons is the horizontal component of the force (very unclear in the question), then the work done is

<u>Work = (force) x (distance)</u>

Work = (53.91 N) x (10 m)

<em>Work = 539.1 Joules</em>

<u>Power = (work done) / (time to do the work)</u>

Power = (539.1 joules) / (90 sec)

Power = (539.1/90) (joule/sec)

<em>Power = 5.99 watts</em>

=====================================

Note:  If the mass of the box is only 2 kg, and you push it along the surface with a constant force of 53⁺ Newtons, and the surface really is frictionless, then that box is gonna cover a WHOOOOOLE LOT more than 10 meters in 90 seconds.  I get 109,168 meters !

5 0
4 years ago
Calculate the gravitational force of the Earth and Moon. The Earth has a mass of 5.972x 1024 kg and the Moon has a mass of 7.348
Wewaii [24]

The gravitational force is 1.98×10^20 m.

Computation of gravitational force:

The gravitational force F between two masses is given by the formula,

F=Gm1m2/ r^2

where G is the gravitational constant.

Note: It is assumed m1=5.972×10^24 kg, m2=7.348×10^22 kg and r=3.84×10^8 m and G=6.67×10^(-11) N m^2 kg^-2

Then F=(6.67×10^(-11)×5.972×10^(24)×7.348×10^(22))/ (3.84×10^8)

         F=1.98×10^(20) N

Check out more about gravitational force here:

brainly.com/question/13054973

#SPJ 4

7 0
2 years ago
Read 2 more answers
A certain tuning fork vibrates at a frequency of 215 Hz while each tip of its two prongs has an amplitude of 0.832 mm. (a) What
Marysya12 [62]

Explanation:

It is given that,

Frequency of vibration, f = 215 Hz

Amplitude, A = 0.832 mm

(a) Let T is the period of this motion. It is given by the following relation as :

T=\dfrac{1}{f}

T=\dfrac{1}{215}

T=4.65\times 10^{-3}\ s

(b) Speed of sound in air, v = 343 m/s

It can be given by :

v=f\times \lambda

\lambda=\dfrac{v}{f}

\lambda=\dfrac{343\ m/s}{215\ Hz}

\lambda=1.59\ m

Hence, this is the required solution.

5 0
4 years ago
A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at 33.0°. It momentarily stops when it has rolled 1.7
galben [10]

Answer:

v = 3.6 m/s

Explanation:

given,

Inclination of the ramp, θ = 33⁰

Distance of the rope, d = 1.70 m

initial speed = ?

using conservation of energy

\dfrac{1}{2}mv^2 + \dfrac{1}{2}I_{com}\omega^2 = m g h

h = d sinθ  

h = 1.7 x sin 33° = 0.926 m

moment of inertia of the solid ball

I_{com}= \dfrac{2}{5}MR^2  

we know, v = r ω

\dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{2}{5}mR^2\times \dfrac{v^2}{R^2}= m g h

\dfrac{1}{2}mv^2 + \dfrac{1}{5}mv^2 = mgh

gh= \dfrac{7}{10}v^2

v = \sqrt{\dfrac{10}{7}gh}

v = \sqrt{\dfrac{10}{7}\times 9.8\times 0.926}

v = 3.6 m/s

Hence, initial speed of the solid ball = 3.6 m/s

8 0
3 years ago
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