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3 years ago

For this graph to the right, what is the velocity of the object? Show your work.

Physics

1 answer:

mr Goodwill [35]3 years ago

7 0

I dont see a mark where it is, if there was a line it would be marking the constant speed of it. There is just time(s) which probably stands for speed, and position(m) which probably stands for miles. There is a measurement of 50, down to 10. Please give me more explanation to this question and I may be able to help!

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user100 [1]

Being hot or not drinking enough fluids.

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4 years ago

Read 2 more answers

Two cars are traveling at the same speed of 27 m/s on a curve thathas a radius of 120 m. Car A has a mass of 1100 kg and car B h

Lera25 [3.4K]

Answer:

$a_{cA} = 6.075$ m/s²

$a_{cB} = 6.075$ m/s²

$F_{cA} = 6682.5 N$

$F_{cB} = 9720 N$

Explanation:

Normal or centripetal acceleration measures change in speed direction over time. Its expression is given by:

$a_{c} = \frac{v^{2} }{r}$ Formula 1

Where:

$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

.(m/s)

r: It is the radius of curvature. (m)

Newton's second law:

∑F = m*a Formula ( 2)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

Data

$v_{A} = 27 \frac{m}{s}$

$v_{B} = 27 \frac{m}{s}$

$m_{A} = 1100 kg$

$m_{B} = 1600 kg$

r= 120 m

Problem development

We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

$F_{cA} = m_{A} *a_{cA} = 1100kg*6.075\frac{m}{s^{2} }$

$F_{cA} = 6682.5 N$

$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

4 0

4 years ago

A student hears a police siren. What would change the frequency that the student hears? check all that apply

andriy [413]

Considering the Doppler efect, the frequency heard by the student would change if:

if the student walked toward the police car.
if the student walked away from the police car.
if the police car moved toward the student.
if the police car moved away from the student.

<h3>Doppler effect</h3>

The Doppler effect is defined as the change in the apparent frequency of a wave produced by the relative motion of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion when the sender and receiver, or observer, move relative to each other.

The following expression is considered the general case of the Doppler effect:

$f'=f\frac{v+-vR}{v-+vE}$

Where:

f', f: Frequency perceived by the receiver and frequency emitted by the transmitter, respectively. Its unit of measurement in the International System (S.I.) is the hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s⁻¹)
v: Wave propagation speed in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 343 m/s.
vR, vE: Receiver and transmitter speed respectively. Its unit of measure in the S.I. is the m/s
±, ∓:
We will use the + sign:

In the numerator if the receiver approaches the sender
In the denominator if the sender moves away from the receiver

We will use the - sign:

In the numerator if the receiver moves away from the sender
In the denominator if the sender approaches the receiver

In summary, the Doppler Effect is an alteration of the observed frequency of a sound due to the movement of the source or the observer, that is, they are changes in the frequency and wavelength of a wave due to the relative movement between the wave source and the observer.

<h3>Changes on the frequency </h3>

In this case, considering the Doppler effect, the frequency heard by the student would change if: