All categories

3 years ago

Why can you see an electric bulb at night when it is switched on ?

Physics

1 answer:

Dominik [7]3 years ago

8 0

Answer:

Explanation:

Because when the electric bulb at night switched It emitted the light this light rays forms an image on our retina hence the bulb become visible at night

2 ans

When an electric bulb is switched on, the electricity is allowed to pass in its circuit and is converted into light energy. Thus, a bulb glows when switched on and the eyes perceives the light energy in the visible spectra which is given out by the bulb. Hence, the bulb becomes visible

You might be interested in

Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you

Alekssandra [29.7K]

Answer:

(a) F = 6.14 *10⁻⁴ N

(b) P = 6.14* 10⁻¹⁰ Pa

Explanation:

Area of sail A = 1.0 km² = 1.0 * 10⁶m²

Wavelength of light λ = 650 nm = 650 * 10⁻⁹ m

Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s

(a)

Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹

= 1.0192 * 10⁻²⁷ kg.m/s

Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.

Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.

dp/dt = R * h/ λ

This is the force acting on the sail.

F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N

F = 6.14 *10⁻⁴ N

Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail

P = F/A = 6.14 * 10⁻⁴ / 1.0 * 10⁶ = 6.14* 10⁻¹⁰ Pa

P = 6.14* 10⁻¹⁰ Pa

Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²

As the spacecraft starts from rest, initial speed u=0,m/s ,

final speed is u = 1.0 m/s after time t

v = u+at

t = 1.0 - 0/ 6.14 * 10⁻⁴ = 1629s = 27.2 min

t = 27.2 min

4 0

3 years ago

Help with 1 2 and 3 please

geniusboy [140]

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. $I = \frac{q}{t}$ Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3. So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

$\frac{p1L1}{A1} = \frac{p2L2}{A2}\\$

We are looking for L2 so we can isolate using algebra to get:

$\frac{A2(\frac{P1L1}{A1}) }{P2} = L2$

If you fill in those values you get 0.0205

or 2.05 cm

6 0

3 years ago

Gravity is a force that helps to hold the universe together.true or false

Valentin [98]

Yes, this is a true statement.

gravity is so important.

7 0

3 years ago

Read 2 more answers

A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t

kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

$\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W$

where we have used the values given by the information of the problem for N B and A.

the emf is given by:

$emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV$

hope this helps!!

5 0

3 years ago

Suppose that a freely falling object were somehow equipped with an odometer. Would the readings of distance fallen each second i

maks197457 [2]

Answer:

1 greater distances fallen in successive seconds

Explanation:

When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement

If we use the kinematic formula we can find the position of the body

Y = Vo t + ½ to t2

Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity

Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2

Let's look for the position for successive times

t (s) Y (m)

1 -4.9

2 -19.6

3 -43.2

The sign indicates that the positive sense is up

It can be clearly seen that the distance is greatly increased every second that passes

3 0

3 years ago