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algol [13]
3 years ago
5

The wavelengths of light emitted by a firefly span the visible spectrum but have maximum intensity near 550 nm. A typical flash

lasts for 100 ms and has a power of 1.2 mW. If we assume that all of the light is emitted at the peak-intensity wavelength of 550 nm, how many photons are emitted in one flash?
Physics
1 answer:
Reil [10]3 years ago
3 0

Answer:

3.3\cdot 10^{15}

Explanation:

First of all, let's calculate the energy of a single photon of wavelength

\lambda=550 nm=5.5\cdot 10^{-7}m

which is given by

E_1 = \frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.5\cdot 10^{-7} m}=3.6\cdot 10^{-19} J

The power of the flash is

P=1.2 mW=0.0012 W

and the time it lasts is

t=100 ms=0.1 s

so the total energy delivered in one flash is

E=Pt=(0.0012 W)(0.1 s)=1.2\cdot 10^{-3}J

This energy contains exactly N photons each of energy E_1, so the number of photons emitted in one flash is

N=\frac{E}{E_1}=\frac{1.2\cdot 10^{-3} W}{3.6\cdot 10^{-19}J}=3.3\cdot 10^{15}

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Your heart pumps 80 g of blood with each beat. The blood starts from rest and reaches a speed of 0.60 m/s in the aorta. If each
Gala2k [10]

Answer:

d. 0.3 N

Explanation:

Force: This is defined as the product of mass of a body and its acceleration.

The S.I unit of Force is Newton (N).

Mathematically Force can be represented as,

F = ma .................. Equation 1

Where F = force, m = mass, a = acceleration.

also

a = (v-u)/t............... Equation 2

Where v = final velocity, u = initial velocity, t = time.

Given: v = 0.60 m/s, u = 0 m/s ( From rest), t= 0.16 s.

Substitute into equation 2

a = (0.60-0)/0.16

a = 3.75 m/s²

Also given: m = 80 g = 0.08 kg.

Substitute into equation 1,

F = 0.08(3.75)

F = 0.3 N.

Hence he average force = 0.3 N

The right option is d. 0.3 N

6 0
3 years ago
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

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