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Over [174]
4 years ago
13

At what distance will a 45 W lightbulb have the same apparent brightness as a 130 W bulb viewed from a distance of 35 m ? (Assum

e that both bulbs convert electrical power to light with the same efficiency, and radiate light uniformly in all directions.)
Physics
1 answer:
Sloan [31]4 years ago
5 0

Answer:

20.6 m

Explanation:

P₁ = Power of first bulb = 45 W

P₂ = Power of second bulb = 130 W

r₁ = distance from the first bulb

r₂ = distance from the second bulb = 35 m

Using the equation

\frac{P_{1}}{r_{1}^{2}} = \frac{P_{2}}{r_{2}^{2}}

Inserting the values

\frac{45}{r_{1}^{2}} = \frac{130}{35^{2}}

r₁ = 20.6 m

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On a playground, there is a merry‑go‑round. In order to get it moving, Bonnie applies a force of 31 N31 N . The merry‑go‑round m
nika2105 [10]

Answer:

The magnitude of the torque is 263.5 N.

Explanation:

Given that,

Applied force = 31 N

Distance from the axis = 8.5 m

She applies her force perpendicularly to a line drawn from the axis of rotation

So, The angle is 90°

We need to calculate the torque

Using formula of torque

\tau=Fd\sin\theta

Where, F = force

d = distance

Put the value into the formula

\tau=31\times8.5\sin90

\tau= 263.5\ N

Hence, The magnitude of the torque is 263.5 N.

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3 years ago
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binge drinking alcohol

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4 years ago
Read 2 more answers
A small economy car (low mass) and a limousine (high mass) are pushed from rest across a parking lot, equal distances with equal
Studentka2010 [4]

Answer:

The car that receives more kinetic energy is the small economy car.

Explanation:

K.E = 0.5*mv²

Where;

K.E is the kinetic Energy

M is the mass of an object

V is the velocity of the moving object

But F = m(v/t), from Newton's second law of motion

If equal forces were applied to the two cars, then the velocity of each car will be calculated as follows.

v = (Ft/m)

v² = (Ft/m)²

Substitute in the value of v² into Kinetic energy equation

K.E = 0.5*mv²

K.E = 0.5*m(Ft/m)² = (0.5*F²t²)/m

Also assuming equal distance, equal force and assuming equal time for both cars.

The above equation will reduce to, K.E = k/m

Where k = 0.5*F²t², which is equal in both cars.

Thus, Kinetic energy will depend only on the mass of each car.

From the above expression, Kinetic Energy received by each car is inversely proportional to the mass of the car.

A small economy car (low mass)  will receive more kinetic energy while a limousine (high mass) car will receive less kinetic energy.

Therefore, the car that receives more kinetic energy is the small economy car.

6 0
3 years ago
Please Help !!!!!!!!!!!!!!!!!!!!!!!!!!
larisa [96]

As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.

Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.

As per Planck's quantum theory -

                                                      hf =E'-E

          Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.

The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.

We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-

       v=\lambda f

[here \lambda is the wavelength of the wave ]

It is given that wavelength =21 cm=0.21 m

 The frequency is calculated as -

                                                         f=\frac{v}{\lambda}

                                                                 =\frac{c}{\lambda}

                                                                 =\frac{3*10^{8} m/s }{0.21 m}

                                                                  =14.2857*10^{8} s^{-1}

 Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz

Here Hertz[ Hz] is the unit of frequency.

7 0
3 years ago
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