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Xelga [282]
4 years ago
12

The linear magnification produced by a spherical mirror is +1/3. Analysing this value. (i) state the type of mirror and (ii) pos

ition of the object with respect to the pole of the mirror. Draw ray diagram to justify your answer.

Physics
1 answer:
elena-s [515]4 years ago
7 0

Answer :

Explanation :

It is given that :

Magnification, m = \frac{+1}{3}

(1 ) Since, the magnification is positive it means it is a convex mirror.

(2)  The image is formed at the back of mirror and the image is virtual and erect.

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A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF. What is the amount of charge stored i
ioda
C = 4 \ \mu F = 4 \cdot 10^{-6} \ F. \newline
q = Cu = 4 \cdot 10^{-6} \cdot 2 = 8 \cdot 10^{-6} = 0.000008 \ C.
8 0
4 years ago
Two positive charged particles will ?
Vika [28.1K]

Answer:

make a negative

Explanation:

yep

6 0
3 years ago
Read 2 more answers
You throw a balloon that floats in the air with a velocity of 2 m / s south . If the wind speed is 5 m / s west , how far south
zvonat [6]

Answer:

The distance traveled by the balloon is 10.77 m

Explanation:

velocity of the ball, v_b = 2 m/s south

velocity of the air, v_a = 5 m/s west

To determine the distance the balloon will travel after 2 seconds, first determine the resultant velocity of the balloon.

                                       | 2m/s

                                       |

                                       |

                                      ↓

      5m/s  ←------------------

the two velocities forms a right angled triangle and the resultant will be the hypotenuses side of the triangle.

R² = 5² + 2²

R² = 29

R = √29

R = 5.385 m/s

The distance traveled by the balloon is calculated as;

d = R x t

where;

t is time of the motion = 2 seconds

d = 5.385 x 2

d = 10.77 m

Therefore, the distance traveled by the balloon is 10.77 m.

4 0
3 years ago
The human ear canal is about 2.3 cm long. If it is regarded as a tube open at one end and closed at the eardrum, what is the fun
mariarad [96]

Answer:

For the given conditions the fundamental frequency is 3728.26 Hertz

Explanation:

We know that for a pipe open at one end and closed at other end the fundamental frequency is given by

f=\frac{V_{s}}{4L}

where

f is the fundamental frequency

V_{s} is the speed of sound in air in the surrounding conditions.

L = Length of the pipe

Applying values we get and using speed of sound as 343m/s we get

f=\frac{343}{4\times 2.3\times 10^{-2}}=3728.26Hz

7 0
3 years ago
A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
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