All categories

3 years ago

an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity

Physics

2 answers:

Elina [12.6K]3 years ago

5 0

Answer:

$displacement = 150 - 65 = 85m \\ time = 5sec \\ velocity = \frac{displacement}{time} \\ = \frac{85}{5} = 17 \frac{m}{s} south \\ thank \: you$

Ilya [14]3 years ago

3 0

Total distance=150m-65m=85m
Time=5s

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Distance}{Total\:Time}$

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}$

$\\ \rm\longrightarrow Avg\:Velocity= 17m/s$

You might be interested in

PLEASE HELP!!! Both Transverse and Longitudinal Waves transmit _______ and not ______.

qwelly [4]

I believe this is it

6 0

4 years ago

PLZ HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!! What is a front?

Shkiper50 [21]

Answer:

in computer and writing context it is many styles of writing that you can make the text look like. I.E: helllo hello

Explanation:

A font is the combination of typeface and other qualities, such as size, pitch, and spacing. For example, Times Roman is a typeface that defines the shape of each character. Within Times Roman, however, there are many fonts to choose from -- different sizes, italic, bold, and so on.

Mark me the brainliest plz

4 0

3 years ago

Read 2 more answers

A particle moves along a path described by y=Ax^3 and x = Bt, where tt is time. What are the units of A and B?

Sergeu [11.5K]

Answer:

In a nutshell, units of A and B are $\frac{1}{[l]^{2}}$ and $\frac{[l]}{[t]}$ , respectively.

Explanation:

From Dimensional Analysis we understand that $x$ and $y$ have length units ( $[l]$ ) and $t$ have time units ( $[t]$ ). Then, we get that:

$[l] = A\cdot [l]^{3}$ (Eq. 1)

$[l] = B\cdot [t]$ (Eq. 2)

Now we finally clear each constant:

$A = \frac{[l]}{[l]^{3}}$

$A = \frac{1}{[l]^{2}}$

$B = \frac{[l]}{[t]}$

In a nutshell, units of A and B are $\frac{1}{[l]^{2}}$ and $\frac{[l]}{[t]}$ , respectively.

4 0

4 years ago

An unknown source plays a pitch of middle C (262 Hz). How fast would the sound wave from this source have to travel to raise

Viefleur [7K]

Answer:

The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

$\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}$

$\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}$

$f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)$

$\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)$

$\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

$v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

Substitute the given values in the formula,

$v_{s}=343+\frac{262}{271}(343-0)$

$v_{s}=343+0.966(343)$

$v_{s}=343-331.33$

$v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Therefore, The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$

4 0

3 years ago

Wanda exerts a constant tension force of 12 N on an essentially massless string to keep a tennis ball (m = 60 g) attached to the

goldfiish [28.3K]

Answer:

r = 0.405m = 40.5cm

Explanation:

In order to calculate the length of the string between Wanda and the ball, you take into account that the tension force is equal to the centripetal force over the ball. So, you can use the following formula:

$F_c=ma_c=m\frac{v^2}{r}$ (1)

Fc: centripetal acceleration (tension force on the string) = 12N

m: mass of the ball = 60g = 0.06kg

r: length of the string = ?

v: linear speed of the ball = 9.0m/s

You solve for r in the equation (1) and replace the values of the other parameters:

$r=\frac{mv^2}{F_c}=\frac{(0.06kg)(9.0m/s)^2}{12N}=0.405m$

The length of the string between Wanda and the ball is 0.405m = 40.5cm

7 0

3 years ago