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2 years ago

What is the formula for finding the velocity?

Physics

2 answers:

Kazeer [188]2 years ago

8 0

Answer:

v = Δs/Δt

and use this calcultor"Velocity Calculator | Definition | Formula - Omni Calculator"if u have problems to slove.cos

Explanation:

blondinia [14]2 years ago

4 0

I hope this answers your question and helps:)

You might be interested in

g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7

bogdanovich [222]

Answer:

When they are approaching each other

$f_a = 2228.7 \ Hz$

When they are passing each other

$f_a = 2100Hz$

When they are retreating from each other

$f_a = 1980.7 Hz$

Explanation:

From the question we are told that

The velocity of car one is $v_1 = 13.0 m/s$

The velocity of car two is $v_2 = 7.22 m/s$

The frequency of sound from car one is $f_e = 2.10 kHz$

Generally the speed of sound at normal temperature is $v = 343 m/s$

Now as the cars move relative to each other doppler effect is created and this can be represented mathematically as

$f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]$

Where $v_s$ is the velocity of the source of sound

$v_o$ is the velocity of the observer of the sound

$f_o$ is the actual frequence

$f_a$ is the apparent frequency

Considering the case when they are approaching each other

$f_a = f_o [\frac{v + v_o}{v - v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 + 7.22}{ 343 - 13} ]$

$f_a = 2228.7 \ Hz$

Considering the case when they are passing each other

At that instant

$v_o = v_s = 0m/s$

$f_o = f_e$

$f_a = f_o [\frac{v }{v } ]$

$f_a = f_o$

Substituting value

$f_a = 2100Hz$

Considering the case when they are retreating from each other

$f_a = f_o [\frac{v - v_o}{v + v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 - 7.22}{343 + 13} ]$

$f_a = 1980.7 Hz$

7 0

3 years ago

Spurs"" are probably the result of ____.

egoroff_w [7]

Spurs are probably the result of <u>self-sustaining</u> <u>star formation.</u>

<h3>What is the formation of gaseous spurs in spiral galaxies?</h3>

The gigantic form of the magnificent doppelganger spiral patterns that spiral outward from the galactic cores gave spiral galaxies their name. These light arms of spiral galaxies are frequently seen in optical pictures to be speckled with bright star-forming areas at regular intervals.

Smaller structures spread forth and rearward into the interarm area from each major spiral arm. Spiral-arm also known as spurs are the name given to these substructures. Sometimes the spurs are also filled with star-forming clusters. As a consequence, we may draw the conclusion that spurs most likely emerge from self-sustaining star formation.

Learn more about the spiral galaxies here:

brainly.com/question/13956361

#SPJ11

5 0

1 year ago

What are 4 physical properties for the bear

Pie

Answer:

A bear normally has a short, thick neck, a rounded head, a pointed muzzle, short ears, and small eyes. Some species have round faces. Bears have poor eyesight, and most have only fair hearing.

Explanation:

Modern Bears are characterized with large body and stocky legs, a long snout, shaggy hair, plantigrade paws with five non-retractile claws and a short tail.

Grizzly bears (Ursus arctos horribilis) have concave faces, a distinctive hump on their shoulders, and long claws about two to four inches long. Both the hump and the claws are traits associated with a grizzly bear's exceptional digging ability. Grizzlies are often dark brown, but can vary from blonde to nearly black.

The brown bear has a slight hump above its shoulder, round ears, a long snout and big paws with long, curved claws that it uses for digging. Unlike the black bear, it can't climb trees. It can weigh between 350-1,500 pounds. When standing on its hind legs it can be up to 5 feet tall.

Hope this helps :)

(I didn't know which type of bear so i did brown bear and grizzly bear)

3 0

3 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$