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2 years ago

13

traveling at about 30 mph, how many feet will the average driver cover from the time they see the danger until they hit the brak

es

1 answer:

barxatty [35]2 years ago

5 0

Answer: 75 ft

Explanation:

Breaking distance = Speed²/ 20

= 30²/20

= 45 feet

Stopping distance = Speed + braking distance

= 30 + 45

= 75 ft

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A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of

Wewaii [24]

(1) The wavelength of the wave is 1.164 m.

(2) The velocity of the wave is 23.7 m/s.

(3) The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

<h3> Wavelength of the wave</h3>

A general wave equation is given as;

y(x, t) = A sin(Kx - ωt)

<h3>Velocity of the wave</h3>

v = ω/K

From the given wave equation, we have,

y(x, t) = 0.048 sin(5.4x - 128t)

v = ω/K

where;

ω corresponds to 128
k corresponds to 5.4

v = 128/5.4

v = 23.7 m/s

<h3>Wavelength of the wave</h3>

λ = 2π/K

λ = (2π)/(5.4)

λ = 1.164 m

<h3>Maximum speed of the wave</h3>

v(max) = Aω

where;

A is amplitude of the wave
ω is angular speed of the wave

v(max) = (0.048)(128)

v(max) = 6.14 m/s

Thus, the wavelength of the wave is 1.164 m.

The velocity of the wave is 23.7 m/s.

The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

Learn more about wavelength here: brainly.com/question/10728818

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3 0

1 year ago

4. State in words how acceleration is calculated.

dimaraw [331]

You multiply force times friction

4 0

2 years ago

Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; from a cert

Brrunno [24]

Answer:

a

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

b

$H = 9.86 \ m$

Explanation:

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

When $v = 50 \ km/h = \frac{50 *1000}{3600} = 13.9 \ m/s$ we have that

$13.9 = \sqrt{ 2 g H}$

=> $H = \frac{13.9^2}{2 * 9.8}$

=> $H = 9.86 \ m$

6 0

2 years ago

What must exist for the engineering process to begin?

Vlad1618 [11]

Answer:

d

Explanation:

rzp-yyib-oiv

plèase jóin ón góógle mèèt

5 0

2 years ago

Read 2 more answers

Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of

Lana71 [14]

Electric field due to a point charge is given as

$E = \frac{kq}{r^2}$

here we know that

$q = 1.6 \times 10^{-19} C$

also the distance is given as

$r = 5.29 \times 10^{-11} m$

now we will have

$E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}$

so we will have

$E = 5.14 \times 10^{11} N/C$

so above is the electric field due to proton

5 0

2 years ago