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Tju [1.3M]
4 years ago
14

A cyclist rides 6.3 km east for 21 minutes, then he turns and heads west for 6 minutes and 1.8 km. Finally, he rides east for 13

.2 km, which takes 44 minutes. (Assume motion toward the east is positive. Indicate the direction with the signs of your answers.) What is the final displacement of the cyclist?
Physics
1 answer:
Leona [35]4 years ago
6 0

Answer:

\vec{d}=17.7km

Explanation:

Displacement is a vector that defines the position of a particle. The vector extends from the initial position to the final position. Therefore, the displacement only takes into account this positions, since its trajectory is not important:

\vec{d}=6.3km-1.8km+13.2km\\\vec{d}=17.7km

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A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be
lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × 10^{-3} m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36  

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ)       ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v  /  2     ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × 10^{-3})²×ω²× 34.64 )   /  2

0.050 ×(6.43 × 10^{-3})²×ω²× 34.64 = 108

ω² = 108 / 7.160  × 10^{-5}

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0
3 years ago
The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?
BigorU [14]

1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

where

c=3\cdot 10^{-8} m/s is the speed of light

\lambda is the wavelength of the wave

In this problem, we have light with wavelength

\lambda=425 nm=425\cdot 10^{-9} m

Substituting into the equation, we find the frequency:

f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


2. Period: 1.42 \cdot 10^{-15}s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s


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snow_tiger [21]
"Constant velocity" is practically a definition for zero acceleration.
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