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3 months ago

9

A grasshopper floating in water generates waves at a rate of three per second with a wavelength of two centimeters. (a) What is

the period of these waves? (b) What is the wave velocity?

1 answer:

Keith_Richards [3.2K]3 months ago

8 0

Answer:

(a) 0.33 seconds

(b) 6 cm/s

Explanation:

The frequency, f = 3 waves per second

and the wavelength, λ = 2 cm = 0.02 m

(a) The wave period is defined as the duration required for one complete oscillation of the wave. This period is the inverse of the frequency.

T = 1 / f = 1 / 3 = 0.33 seconds

(b) The relationship between wave speed, frequency, and wavelength is determined by

v = f x λ

v = 3 x 0.02 = 0.06 m/s

v = 6 cm/s

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A ball with a mass of 0.5 kilograms is lifted to a height of 2.0 meters and dropped. It bounces back to a height of 1.8 meters.

kicyunya [3294]

Hello! Thanks for sharing your query here.

To determine the change in potential energy, you would utilize the formula:

delta PE = mg*delta h
delta PE = 0.5*9.81*(2-1.8)
delta PE = 0.98 J

The kinetic energy is derived from the potential energy.

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2 months ago

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A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What

kicyunya [3294]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = $60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s$

s = Distance traveled = $123\ ft=\dfrac{123}{3.281}=37.4885705578\ m$

$\theta$ = Incline = $26^{\circ}$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2$

Friction coefficient

$\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561$

$mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m$

The calculated stopping distance is 83.1946504051 m

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4 months ago

A brass lid screws tightly onto a glass jar at 20 degrees C. To help open the jar, it can be placed into a bath of hot water. Af

serg [3582]

Answer:

0.0016 cm

Explanation:

$\alpha_b$ = The thermal expansion coefficient for brass = $19\times 10^{-6}\ /^{\circ}C$

$\alpha_g$ = The thermal expansion coefficient for glass = $9\times 10^{-6}\ /^{\circ}C$

$\Delta T$ = Temperature change = $(60-20)^{\circ}C$

$R_0$ = Original radius = 4 cm

The change in radius is described by

$R=R_0(1+\alpha\Delta T)$

The radius difference between the lid and the jar

$\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm$

The gap measures 0.0016 cm or 0.000016 m

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3 months ago

You're driving your pickup truck around a curve that has a radius of 22 m.How fast can you drive around this curve before a stee

Sav [3153]

The maximum speed around this curve before a toolbox made of steel slips off the truck bed is: 14.832 m/s

Further explanation

The centripetal force acts on objects in circular motion, directed towards the circle's center.

$\large {\boxed {\bold {F = \frac {mv ^ 2} {R}}}$

F = centripetal force, N

m = mass, Kg

v = linear velocity, m / s

r = radius, m

The velocity directed toward the center of the circle is referred to as linear velocity.

It can be represented as:

$\displaysyle v = 2 \pi.r.f$

r = radius of the circle

f = rotations per second (RPS)

Pickup trucks negotiating a curve are influenced by centripetal forces. To prevent the steel toolbox from slipping off the bed of the truck, the centripetal force acting on it must equal its weight. Should the centripetal force surpass the toolbox's weight, it will fall off.

centripetal force = weight

$\rm \dfrac{mv^2}{r}=mg$

Thus, the maximum velocity to keep the toolbox secure is:

$\rm v^2=r\times g\\\\v=\sqrt{r\times g}$

For a curve with a radius of 22 m, it follows that:

$\rm v=\sqrt{22\times 10}\\\\v=\sqrt{220}\\\\v=\boxed{\bold{14.832\:m/s}}}$

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