All categories

2 years ago

A=vf-vi/t solve for vi

1 answer:

7 0

A = vf - vi/t
A - vf = -vi/t Multiply both sides by -1
-A + vf = vi/t
vf - A = vi/t Cross multiply both sides by t.
tvf - At = vi

vi = tvf - At

You might be interested in

Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

seraphim [82]

Answer:

$\theta = 15^o \: or\: 75^o$

Explanation:

As we know that the formula of range is given as

$R = \frac{v^2sin2\theta}{g}$

now we know that

maximum value of the range of the projectile is given as

$R_{max} = \frac{v^2}{g}$

now we need to find such angles for which the range is half the maximum value

so we will have

$\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}$

$sin(2\theta) = \frac{1}{2}$

$2\theta = 30 or 150$

$\theta = 15^o \: or\: 75^o$

7 0

2 years ago

Please answer I will give good rating and mark brainlest

lina2011 [118]

The answer is C. You must divide your wavelength and your frequency to get your answer.

8 0

2 years ago

a hunter 412.5m from a cliff moves a distance x towards the cliff and fires a gun. he hears the echo from the cliff after 2.2sec

Inessa [10]

Answer: 49.5 m

Explanation:

The speed of sound $s$ is given by a relation between the distance $d$ and the time $t$ :

$s=\frac{d}{t}$ (1)

Where:

$s=330 m/s$ is the speed of sound in air (taking into account this value may vary according to the medium the sound wave travels)

$d=412.5 m-x$ since we are told th hunter was initially 412.5 meters from the cliff and then moves a distance $x$ towards the cliff

$t=\frac{2.2 s}{2}=1.1 s$ Since the time given as data (2.2 s) is the time it takes to the sound wave to travel from the hunter's gun and then go back to the position where the hunter is after being reflected by the cliff

Having this information clarified, let's isolate $d$ and then find $x$ :

$d=st$ (2)

$412.5 m-x=(330 m/s)(1.1 s)$ (3)

Finding $x$ :

$x=49.5 m$ This is the distance at which the hunter is from the cliff.

3 0

2 years ago

A charge q produces an electric field of strength 2E at a distance of d away. Determine the electric field strength at a distanc

Alja [10]

The electric field strength of a point charge is inversely proportional to the square of the distance from the charge ... a lot like gravity.

If the magnitude of the field is (2E) at the distance 'd', then at the distance '2d', it'll be (2E)/(2²). That's (2E)/4 = 0.5E .

3 0

2 years ago

Read 2 more answers

What happens to light waves from a star as the star moves away from Earth?

AURORKA [14]

<h2>Answer: Light waves have a redshift due to the Doppler effect </h2>

The astronomer Edwin Powell Hubble observed several celestial bodies, and when obtaining the spectra of distant galaxies he observed the spectral lines were displaced towards the red (red shift), whereas the nearby galaxies showed a spectrum displaced to the blue.

From there, Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum. <u>The same happens with the stars and this phenomenom is known as the Doppler effect. </u>

This phenomenon refers to the change in a wave perceived frequency (or wavelength=color) when the emitter of the waves, and the receiver (or observer in the case of light) move relative to each other. For example, as a star moves away from the Earth, its espectrum turns towards the red.

8 0

2 years ago