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3 years ago

What is the difference between P and S waves

Physics

1 answer:

Doss [256]3 years ago

3 0

Primary waves are faster than Secondary waves

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Isabela is an astronomer studying stars A and B. Star A is farther away from Isabela and yet when Isabela observes the stars, th

kakasveta [241]

Answer: star A is more luminous

Explanation:

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3 years ago

I NEED HELP ASAP WILL MARK BRAINLIEST:

Evgen [1.6K]

Answer:

It is b for sure.

Explanation:

because they are examining urine now that's pure science.

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3 years ago

A ____ is one of the tiny dots of light that form a grid on your screen.

Degger [83]

Hi there

The answer is pixel

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2 years ago

Read 2 more answers

What is the speed of a wave on a string with a wavelength of 1.75 m and a frequency of 2.0 Hz

deff fn [24]

Answer:

V=3.5 m/s

Explanation:

V=(F)(W)

V=(2)(1.75)

V= 3.5 m/s

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2 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

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3 years ago