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alekssr [168]
3 years ago
7

i) A 100 W and 60 W bulb are joined in series and connected to the mains. Which bulb will glow brighter? Why? ii) A 100 W and a

60 W bulb are joined in parallel to the mains. Which bulb will glow brighter? Why?
Physics
1 answer:
Reika [66]3 years ago
7 0

Answer:

i) 60 W

ii) 100 W

Explanation:

In each case, the bulb that dissipates the most power is the bulb that glows brighter.  Power is voltage times current (P = VI).  Using Ohm's law, we can rewrite this as P = I²R or P = V²/R.

Bulbs are rated at a certain power for a certain voltage.  P = V²/R, so the bulb with the lower resistance will have the higher power rating.  Therefore, the 100 W bulb has a lower resistance than the 60 W bulb.

i) They are in series, so they have the same current.  P = I²R, so the bulb with the higher resistance will glow brighter.  That's the 60 W bulb.

ii) They are in parallel, so they have the same voltage.  P = V²/R, so the bulb with the lower resistance will glow brighter.  That's the 100 W bulb.

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Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude of the tot
guapka [62]

Answer:

F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

R = \frac{d}{\sqrt2}

now we have

E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}

so the force on the charge is given as

F = QE

F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}

3 0
3 years ago
An electromagnetic wave of wavelength
Ivanshal [37]

Answer:

4.01\cdot 10^{-7} m

Explanation:

When an electromagnetic wave passes through the interface between two mediums, it undergoes refraction, which means that it bents and its speed and its wavelength change.

In particular, the wavelength of an electromagnetic wave in a certain medium is related to the index of refraction of the medium by:

\lambda=\frac{\lambda_0}{n}

where

\lambda_0 is the wavelength in a vacuum (air is a good approximation of vacuum)

n is the refractive index of the medium

In this problem:

\lambda_0 = 5.89\cdot 10^{-7} m is the original wavelength of the wave

n = 1.47 is the index of refraction of corn oil

Therefore, the wavelength of the electromagnetic wave in corn oil is:

\lambda=\frac{5.89\cdot 10^{-7}}{1.47}=4.01\cdot 10^{-7} m

8 0
3 years ago
A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50
laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

  • Height of the window=1.5 m
  • Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

s=ut+\dfrac{at^2}{2}

Let u be the velocity of the ball when it reaches the top of the window

then

1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

3.96=gt\\t=0.4\ \rm s\\

Let h be the height above the top of the window

h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m

3 0
3 years ago
(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r
Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface U_1=\frac{GM_em}{R_e}

Gravitational Potential Energy at height h is U_2=\frac{GM_em}{R_e+h}

Energy required to lift the satellite E_1=U_1-U_2

E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

h=\frac{R_e}{2}

h=3.19\times 10^6\ m

(b)For greater height E_1  is greater than E_2

thus energy to lift the satellite is more than orbiting around earth

4 0
3 years ago
Evaporation only occurs at the surface of a liquid
Veseljchak [2.6K]

yes

evaporation starts on the surface

8 0
4 years ago
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