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iren2701 [21]
2 years ago
15

The electrical voltage between the ends of a resistor is equal to 8 V. A student has two voltmeters available that measure up to

6 V. The voltages of the voltmeters are equal to 500 ohms and 600 ohms, respectively. What voltages will the connected voltmeters indicate? (Fig. 3.45)​

Physics
1 answer:
Rufina [12.5K]2 years ago
6 0

Answer:

I = V / R

I1 = 6 / 500 = .012   amps for full scale deflection

I2 = 6 / 600 = .01   amps for full scale deflection

I = 8 / 1100 = .00727   amps thru load (actual current)

D1 = .00727 / .012 = .606 of full scale deflection

D2 = .00727 / .01 = .727    of full scale deflection

V1 = .606 * 6 = 3.64 V  (of 6 V)

V2 = .727 * 6 = 4.36 V   (of 6 V)

Check: V1 + V2 = 3.64 + 4.36 = 8 V

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An object of mass 'm' is on an inclined plane with an
Levart [38]

Answer:

F_{\text{par}} = F_{\text{frict}}

F_{\text{norm}} = F_{\text{perp}}

Explanation:

See attachment for complete work.

Download pdf
7 0
2 years ago
In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int
Aleonysh [2.5K]

Answer:

a) <em>8.33 x 10^-6 Pa</em>

b) <em>8.23 x 10^-11 atm</em>

c) <em>1.67 x 10^-5 Pa</em>

d) <em>1.65 x 10^-10 atm</em>

<em></em>

Explanation:

Intensity of the light I = 2500 W/m^2

speed of light c<u> </u>= 3 x 10^8 m/s

a) we know that the pressure for for a totally absorbing surface is given as

P_{abs} = I/c = 2500/(3 x 10^8) = <em>8.33 x 10^-6 Pa</em>

b) 1 atm = 101325 Pa

P_{abs} = (8.33 x 10^-6)/101325 = <em>8.23 x 10^-11 atm</em>

c) for a totally reflecting surface

P_{ref} = 2I/c = twice the value for totally absorbing

P_{ref}  = 2 x 8.33 x 10^-6 = <em>1.67 x 10^-5 Pa</em>

d)  1 atm = 101325 Pa

P_{ref} = 2 x 8.23 x 10^-11  = <em>1.65 x 10^-10 atm</em>

8 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
matrenka [14]

the answer is d it reflects all the wavelengths of visible light.

4 0
3 years ago
Read 2 more answers
Statement A: 2.567 km, to two significant figures. Statement B: 2.567 km, to three significant figures. Determine the correct re
sammy [17]

Answer:

Statement A is greater than Statement B.

Explanation:

Statement A: 2.567 km, to two significant figures..

To 2 sig figures means only 2 whole numbers should be left after approximation. Thus, 2.567 to 2 significant figures is 2.6 km

Statement B: 2.567 km, to three significant figures. To 3 sig figures means only 3 whole numbers should be left after approximation. Thus, 2.567 to 3 significant figures is 2.57 km

Comparing both values, statement A is obviously greater than Statement B

5 0
3 years ago
The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
3 years ago
Read 2 more answers
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