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ira [324]
2 years ago
10

Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi

ll perceive the liquid to be what apparent depth?
Physics
1 answer:
Tju [1.3M]2 years ago
7 0

By law of refraction we know that image position and object positions are related to each other by following relation

\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}

here we know that

\mu_1 = 1.473

h_o = 5 cm

\mu_2 = 1

now by above formula

\frac{1.473}{5} = \frac{1}{h_i}

h_i = 3.39 cm

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

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A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee
slamgirl [31]

Answer:

i = 4.9 A

Explanation:

Force on a current carrying rod due to magnetic field is given as

F = iLB

here we know that

i =current in the rod

B = 0.10 T

L = 1.0 m

now magnetic force is balanced by the weight of the rod

so we will have

iLB = mg

i(1.0)(0.10) = 0.05 \times 9.8

i = 4.9 A

8 0
2 years ago
A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal
PilotLPTM [1.2K]
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline 
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
3 0
2 years ago
It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applica
denis23 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is  \theta =   24.2^o

Explanation:

From the question we are told that

     The length of eire let out is  L = 250 \ m

      The emf generated is \epsilon = 40 V

      The earth magnetic field is B = 5.0 *10^{-5} T

     The speed of the shuttle and tether is v =  7.80 * 10^3 \  m/s

The emf generated is mathematically represented as

                             \epsilon = L\ v\ B\ sin \ \theta

making \theta  the subject of the formula

                        \theta =   sin ^{-1}[ \frac{\epsilon}{L  * B  *v} ]

substituting values

                        \theta =   sin ^{-1}[ \frac{40}{250  * (5*10^{-5})  *(7.80 *10^{3})} ]

                        \theta =   24.2^o

6 0
2 years ago
A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18
Annette [7]

Answer:

Part a)

Induced EMF when length vector is along Z direction is 0.72 V

Part b)

Induced EMF when length vector is along Y direction is ZERO

Explanation:

As we know that the motional EMF induced in the wire is given as

E = (v\times B). L

1)

As we know that

v = 18\hat i + 24\hat j + 72\hat k

B = 0.080 \hat j

L = 0.50 \hat k

now we have

\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

E = 1.44 (0.50) = 0.72 V

2)

If the length vector is along Y direction then we have

L = 0.50 \hat j

so again we have

\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

EMF = 0

6 0
2 years ago
A sample of an unknown volatile liquid was injected into a Dumas flask (mflask = 27.0928 g, Vflask = 0.1040 L) and heated until
NNADVOKAT [17]

Answer:

The gas was Hexane

Explanation:

taking the diference between the mass of the flask and the final mass qe can calculate the mass of liquid injected (assuming none escaped the flask):

m_{l}  = 27.4593g - 27.0928g = 0.3665g

with the volume of the flask we can get the density of the gas at the indicated pressure and temperature:

d_{g}  = \frac{0.3665 g}{0.1040L} = 3.524 g/L

From the ideal gases law we have that the density can be calculated as:

d_{g}  = \frac{P*M}{R*T}

Where R is the ideal gases constant = , and M the molecular weight of the fluid. Solving for M:

M=\frac{d_{g}*R*T}{P}=\frac{3.524g/L*0,082atmL/molK*291K}{0.976atm}

M=86.16 g/mol

Note that the temperature is computed in Kelvin T= 18+273=291K

The gas with the closer molar mass is Hexane

4 0
2 years ago
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