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3 years ago

6

A parachute on a racing dragster opens and changes the speed of the car from 95 m/s to 35 m/s in a period of 6.5 seconds. What i

s the acceleration of the dragster?

1 answer:

tatiyna3 years ago

6 0

Answer:

30 miles

Explanation:

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The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

3 years ago

A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed

oksian1 [2.3K]

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

$v=30j$

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

$u=475i$

c. The true velocity of the jet as a vector will be represented as w:

$w=u+v$

$w=475i+30j$

d. The true speed of the jet will be calculated as:

$IwI=\sqrt{(475)^2+(30)^2}$

$IwI=\sqrt{225625+900}$

$IwI=\sqrt{226525}$

$IwI=476 mi/h$

e. The direction of the jet will be:

$tita=tan^{-1}\frac{30}{475}$

$tita=tan^{-1}(0.0632)$

$tita=3.62degrees,or,N86.38degreesS$

7 0

2 years ago

Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57

Harlamova29_29 [7]

We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
$n_1sin(\theta_1)=n_2sin(\theta_2)$
Where $\theta_2$ differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, $\Delta x$ is the distance between both rays.
$\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705$
$\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21$
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
$d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m$
For red we have:
$d_{red}=h.tan(\theta_{2red})\approx 0.0134m$
We finally have:
$\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m$

6 0

2 years ago

If the distance to a distant galaxy is recorded as 10,000 light years, then light must have been traveling for 5000 years to rea

satela [25.4K]

False false false false

5 0

2 years ago

Read 2 more answers

What is net force?

kramer

Answer:

A. The sum of all the forces acting on an object.

3 0

2 years ago