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Ksivusya [100]
3 years ago
6

A car’s 30.0-kg front tire is suspended by a spring with spring constant k=1.00x10^5 N/m. At what speed is the car moving if was

hboard bumps on the road every 0.750 m drive the tire into a resonant oscillation?
Physics
1 answer:
Taya2010 [7]3 years ago
6 0

we know the equation for the period of oscillation in SHM is as follows:

T = 2 * pi * sqrt(mass/k)

we know f = 1/T, so f = 1/(2 * pi) * sqrt(k/m).

since d = v*T, we can say v = d/t = d * f

the final equation, after combining everything, is as follows:

v = d/(2 * pi) * sqrt(k/m)

by plugging everything in

v = .75/(2 * pi) * sqrt((1 * 10^5)/(30))

We find our velocity to be:

v = 6.89 m/s

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Both the lens and the cornea of the eye have a primary function of
3241004551 [841]

Answer: B. bending light

Explanation:

The phenomenom of vision in human eye is thanks to refraction (when light changes its direction as it passes through one medium to another), and this is what the cornea and the lens do.

When the ray of light encounters the eye, the first thing it finds is the <u>cornea</u>, which<u> bends this ray and begins to form an image</u>, then light passes through the <u>pupil</u>, which is in charge of regulating the amount of light that enters in the eye.  

After light travels through pupil it passes through the <u>lens</u>, where <u>the rays of light change the direction again in order to focus the formed image on the retina. </u>

At this point it is important to note the formed image is downward, then the retina transforms light into electrical impulses that are sent to the brain through the optic nerve and finally the brain interprets these messages, and forms a right upward image.

In the image attached these parts can be seen.

6 0
3 years ago
Waves A and B have the same speed, but wave A has a shorter wavelength. Which wave has the higher frequency?
Nuetrik [128]
Frequency and wavelength are inversely proportional.
A shorter wavelength implies a higher frequency.
3 0
3 years ago
An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11
Arada [10]

Answer:

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

8 0
3 years ago
Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm
cestrela7 [59]

Answer:

0.043 m

Explanation:

From the attachment, the shaded part is the ethyl alcohol. The crossed part on the other hand, is that of glycerin.

The height of the Ethyl Alcohol is h2 = 0.25 m, it's density is ρ2 = 790 kg/m³. The density of glycerin is ρ1 = 1260 kg/m³

If we assume pressure at the two points to be the same, then

P1 = P2

ρ1.g.V1 = ρ2.g.V2

ρ1.A.h1 = ρ2.A.h2

ρ1.h1 = ρ2.h2, making h1 subject of formula

h1 = ρ2.h2 / ρ1, so that

h1 = 790 * 0.25 / 1260

h1 = 197.5 / 1260

h1 = 0.157 m

Δh = 0.2 - 0.157

Δh = 0.043 m or 4.3 cm

8 0
3 years ago
7. An engineer is using a wire that has a resistance of 1.5. This resistance is too high for the application he is designing. Th
Nataliya [291]

Answer:

Explanation:

The resistance of a wire depends on the length of the wire and the area of crossection of wire.

resistance is directly proportional to the length of the wire

and inversely proportional to the area of crossection of the wire.

So, to decrease the resistance, length should be decreased and the area of crossection should be increased.

8 0
3 years ago
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