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3 years ago

A truck is traveling 20m/s accelerates 3 m/s^2 for 4 seconds how far did it travel while it was accelerating. Using guess method

Physics

1 answer:

stellarik [79]3 years ago

5 0

<h3> $\huge\underline\bold\blue{ƛƝƧƜЄƦ}$ </h3><h3>Given</h3>

$\blue\star$ v = 20m\s

$\blue\star$ a = 3m\s^2

$\blue\star$ t = 4sec

Firstly we have to find u

$\star$ a = $\dfrac{v - u}{t}$

$\star$ 3m\s = $\dfrac{20 - u}{4}$

$\star$ 12m\s = 20 - u

$\star$ 20 - u = 12m\s

$\star$ - u = -8

$\star$ u = 8

Now we can easily find distance by using second equation of motion

$\red\star$ s = ut + 1\2 at^2

$\red\star$ s = 8(4) + 1\2(3)(16)

$\red\star$ s = 32 + 24

$\red\star$ s = 56

So distance is 56 m\s hope it helps

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Which galaxy is the most stretched out?<br>

RideAnS [48]

Answer:

this is the anwser

Explanation:

The oddball spiral galaxy, called Messier 66, is one-thirdof the Leo Triplet, a group of three interacting galaxies about 35 millionlight-years from Earth (a light-year is the distance light can cover in ayear).

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3 years ago

Read 2 more answers

A roller coaster car is going over the top of a 15-m-radius circular rise. The passenger in the roller coaster has a true weight

konstantin123 [22]

Answer:

v = 7.67 m/s

Explanation:

The equation for apparent weight in the situation of weightlessness is given as:

Apparent Weight = m(g - a)

where,

Apparent Weight = 360 N

m = mass passenger = 61.2 kg

a = acceleration of roller coaster

g = acceleration due to gravity = 9.8 m/s²

Therefore,

360 N = (61.2 kg)(9.8 m/s² - a)

9.8 m/s² - a = 360 N/61.2 kg

a = 9.8 m/s² - 5.88 m/s²

a = 3.92 m/s²

Since, this acceleration is due to the change in direction of velocity on a circular path. Therefore, it can b represented by centripetal acceleration and its formula is given as:

a = v²/r

where,

a = centripetal acceleration = 3.92 m/s²

v = speed of roller coaster = ?

r = radius of circular rise = 15 m

Therefore,

3.92 m/s² = v²/15 m

v² = (3.92 m.s²)(15 m)

v = √(58.8 m²/s²)

7 0

3 years ago

I NEED HELP!!!!!!!!!!!

nadezda [96]

1. 168.1 Hz

To find the apparent frequency heard by the driver in the car, we can use the formula for the Doppler effect:

$f'=(\frac{v\pm v_o}{v\pm v_s})f$

where

f is the original sound of the horn

v is the speed of sound

$v_o$ is the velocity of the observer (the driver and the car), which is positive if the observer is moving towards the source and negative if it is moving away

$v_s$ is the velocity of the sound source (the train), which is positive if the source is moving away from the observer and negative otherwise

In this problem we have, according to the sign convention used:

$v = 343 m/s\\f = 164 Hz\\v_o = -15 m/s\\v_s = -23 m/s$

Substituting, we find:

$f'=(\frac{343-15}{343-23})(164)=168.1 Hz$

2. $2.96\cdot 10^8 m/s$

The speed of light can be calculated as

$v=\frac{d}{t}$

where

d is the distance travelled

t is the time taken

In this problem:

$d=2\cdot 3.85\cdot 10^8 =7.7\cdot 10^8 m$ is the total distance travelled by the laser beam (twice the distance between the Earth and the Moon)

t = 2.60 s is the time taken

Substituting in the formula,

$v=\frac{7.7\cdot 10^8 m}{2.60 s}=2.96\cdot 10^8 m/s$

6 0

3 years ago

A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r

katen-ka-za [31]

Answer:

k = 167.33 N/m

Explanation:

The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
v = c = λ*f

where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =

4.92*10⁷ m.

Solving for f, we get the frequency of the radio waves:

f = 6.1 Hz

Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by a fixed relationship between the spring constant k and the mass m, as follows:

$\omega_{o}^{2} =\frac{k}{m} (1)$

Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

$\omega = 2*\pi *f (2)$

We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
$k = 4*\pi ^{2}*f^{2} *m = 4*\pi ^{2} * (6.1Hz)^{2} * 0.114 kg = 167.33 N/m$

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3 years ago

Ultrasonic images are obtained from the inside organs of our body. This process uses which property of sound wave?

Temka [501]

This question involves the concepts of echo, ultrasonic images, ultrasonic sound waves.

The process of ultrasonic images uses the "echo" property of the sound waves.

Echo is the property of the sound wave by the virtue of which the sound wave reflects back to the source of the sound after hitting a surface or an object.

Ultrasonic images are obtained from inside organs of our body. This process involves the use of ultrasonic sound waves that have a frequency greater than 20,000 Hz. These sound waves are out of the range of audible sound by the human ear. When these ultrasonic sound waves are sent in form of pulses into the human body by the use of probes, they reflect back from the tissues of different organs to the probe. The probe then records the reflection properties of these sound waves and displays them in form of an image, known as ultrasonic images.

Learn more about echo here:

brainly.com/question/14335186?referrer=searchResults

The attached picture shows the process of ultrasonic imaging.

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2 years ago