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3 years ago

Calculate the acceleration of a bus whose speed changes from 7 m/s to 16 m/s over a period of 5 s.

1 answer:

7 0

Let’s do this together!

Okay so the acceleration formula is vf-vi over time .

So the initial velocity (vi) 7m/s final velocity (vf) is 16m/s so we’re going to subtract 16-7 which is 9
M/s

So the time is 5s so 9m/s divided into 5s is 1.8m/s/2

So the answer is 1.8m\s2

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A photon of red light (wavelength = 710 nm) and a ping-pong ball (mass = 2. 40 × 10^-3 kg) have the same momentum. At what speed

Paha777 [63]

The speed of the ball moving is

$v = 3.94 \times 10 {}^{ - 25}m/s$

what is momentum?

The momentum p of a classical object of mass m and velocity v is given by pclassical =mv.

For photons with wavelength λ,this equation does not hold.Instead, the momentum of the Photon is given by p Photon = h/λ

where,h is the planck's constant.

The momentum of the red Photon is

given:

$h = 6.626 \times 10 {}^{ - 34}kgm {}^{2}/s(plancks \: constant)$

$λ = 700 \times 10 {}^{ - 9} m(photons \: wavelength)$

$p \: photons = \frac{6.626 \times 10 {}^{ - 34}kgm {} ^{2}/s }{700 \times 10 {}^{ - 9}m }$

$= 9.47 \times 10 {}^{ - 28}kgm /s$

since,the Photon and the ping-pong ball have the same momentum,we have

$pball = pphotons = \frac{6.626 \times 10 {}^{ - 34}kgm/s }{700 \times 10 {}^{ - 9} }$

$pball = mv,m = 2.40 \times 10 {}^{ - 3}kg$

$v = 3.94 \times 10 {}^{ - 25} m/s$

Therefore, if the red photon and the ping-pong ball have the same momentum, the ping-pong ball must have a speed of approximately

$v = 3.94 \times 10 {}^{ - 25}m/s$

learn more about momentum of photon from here: brainly.com/question/28197406

#SPJ4

3 0

2 years ago

The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70

MrRa [10]

Answer:

a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

C = $\epsilon_o \frac{A}{d}$

C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

C = $\frac{Q}{\Delta V}$

Q₀ = C ΔV

Q₀ = 2.62 10⁻¹² 8.70

Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

C₁ = 1.04 10⁻¹² F

C₁ = Q₀ / ΔV₁

ΔV₁ = Q₀ / C₁

ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

ΔV₁ = 21.9 V

b) initial stored energy

U₀ = $\frac{Q_o}{ 2C}$

U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)

U₀ = 99.2 10⁻¹² J

c) final stored energy

U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)

U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

W = U_f - U₀

W = (249.2 - 99.2) 10⁻¹²

W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0

3 years ago

Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on

Zolol [24]

Answer:71 dB

Explanation:

Given

sound Level $\beta _1=124 dB$

distance $r_1=5.01 m$

From sound Intensity

$\beta =10dB\log (\frac{I_1}{I_0})$

$124=10dB\log (\frac{I_1}{I_0})$

$12.4=\log (\frac{I_1}{I_0})$

$I_1=(1\times 10^{-12})\times 10^{12.4}$

$I_1=2.51 W/m^2$

we know Intensity $I\propto ^\frac{1}{r^2}$

$I_1r_1^2=I_2r_2^2$

$I_2=I_1(\frac{r_1}{r_2})^2$

$I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2$

$I_2=1.24\times 10^{-5} W/m^2$

Sound level corresponding to $I_2$

$\beta _2=10\log (\frac{I_2}{I_0})$

$\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})$

$\beta _2=70.93\approx 71 dB$

6 0

3 years ago

A heavy rope, 60 ft long, weighs 0.7 lb/ft and hangs over the edge of a building 140 ft high. (a) How much work W is done in pul

Marina CMI [18]

Answer:

5880lb-ft of work is done

Explanation:

The length of the heavy rope is given as 60ft and the weight per length is 0.7lb/ft.

Therefore, the total weight of the heavy rope is

60×0.7 =42lb.

The work done in pulling the heavy rope to the top of the building is w = Fd.

Where

F is force is measured in pounds;42lb

d is distance through which the heavy rope is to be pulled measured in feet; 140ft

w= 42lb×140ft= 5880lb-ft

4 0

3 years ago

How does a compound eye work?

Reil [10]

<span>The Compound Eye. The arthropod (e.g., insects, crustaceans) eye is built quite differently from the vertebrate eye (and the mollusk eye). Arthropod eyes are called compound eyes because they are made up of repeating units, the ommatidia, each of which functions as a separate visual receptor.</span>

8 0

3 years ago