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3 months ago

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An engine performs 2700 J of work on a scooter. The scooter and rider have a combined mass of 150 kg. If the scooter started at

rest, what is the speed of the bike after the work is performed?

1 answer:

serg [3.5K]3 months ago

6 0

According to the principle of energy conservation, the engine's work in moving the scooter is converted into the scooter's kinetic energy, represented as:

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Two golf carts have horns that emit sound with a frequency of 394 Hz. The golf carts are traveling toward one another, each trav

Sav [3153]

The frequency detected is 394 Hz. This question pertains to the Doppler effect, outlined by the equation fo = {c + vo}/{c - vs} × f. Here, fo is the observed frequency, c denotes sound speed at 345 m/s, vo is the observer's velocity of 9.5 m/s, and vs refers to the source's velocity of -9.5 m/s (the negative indicates opposite directions). The source's frequency is given as 394 Hz. Substituting the values leads to fo = {345 + 9.5}/{345 + 9.5} × 394. Simplifying yields fo = (354.5/354.5) × 394 = 1 × 394 = 394 Hz.

6 0

2 months ago

If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100

serg [3582]

Answer:

The charge that enters a segment of axon measuring 0.100 mm is $8.98\times 10^{-12} C$

Explanation:

The electric field E produced at a certain point by a point charge is expressed as

$E=k \frac{q}{r^2}$

where $k$ represents the constant = $9.0 \times 10^9 Nm^2 / C^2$

$q$ denotes the point charge's magnitude, and $r$ is the distance from the point charge

The amount of charge entering one meter of the axon equals $5.\times 10^{11} \times (+e)$

The charge that enters a 0.100 mm length of the axon is $5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}$

by substituting the value of $+e=1.6\times 10^{-19} C$ into the equation above, we find that the charge entering a 0.100 mm segment of the axon is

$q=5.\times 10^{11} \times1.6\times 10^{-19} \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C$

3 0

2 months ago

Calculate the wavelength of a photon having 3.26 x 10^-19 joules of energy

Keith_Richards [3271]

The energy contained in a photon is determined by the formula:
$E=hf$
where h represents the Planck constant and f signifies the frequency of the photon. Given the energy of the photon, $E=3.26 \cdot 10^{-19} J$ , we can rearrange the equation to deduce the photon's frequency:
$f= \frac{E}{h}= \frac{3.26 \cdot 10^{-19}J}{6.6 \cdot 10^{-34}Js}=4.94 \cdot 10^{14}Hz$

Now, we can use the relationship that links frequency f, wavelength $\lambda$ , and the speed of light c to ascertain the wavelength of the photon:
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{4.94 \cdot 10^{14} Hz}=6.07\cdot 10^{-7} m=607 nm$

8 0

4 months ago

In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Yuliya22 [3333]

Response:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Components of velocity

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Components of acceleration

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

The time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle covers on the ground is 0.3677181864 m

6 0

2 months ago