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3 years ago

Use the graph below to answer the following question: What is happening to the object's velocity?

Physics

1 answer:

hjlf3 years ago

4 0

The object's velocity is decreasing.

Explanation:

From graph is the attached image, we can clearly point that the velocity of this motion is decreasing with time.

Velocity is a vector quantity.

The y-axis represent displacement.
The x-axis depicts time
Using the graph, we know that the slope of the line on the graph gives us the velocity as it denotes the change of displacement with time.
When we find the slope, it will give us a negative value which shows that the body is slowing down and not increasing speed.

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

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A belief, judgment, or way of thinking about something is called a(n) __________.

NISA [10]

That is an opinion. Opinion is not fact, although it can be based upon fact.

8 0

3 years ago

Read 2 more answers

Does the u declined adverb express time, place, or manner?<br> The squirll spotted us and ran away.

AVprozaik [17]

I feel like it expresses time, but if there is another answer and they say otherwise, then they are probably right.
Hope this helps!

3 0

3 years ago

A cyclotron with a magnetic field of 0.47 t is designed to accelerate protons in a circle with a radius 0.68 m. What is the angu

Ghella [55]

The angular frequency of the cyclotron is 0.07 x $10^{8}$ Hz.

<h3>What is angular frequency?</h3>

Angular frequency, abbreviated "ω" is a scalar measure of rotation rate in physics.
It describes the rate of change of the argument of the sine function, the rate of change of the phase of a sinusoidal waveform, or the angular displacement per unit of time.

<h3>What is cyclotron?</h3>

The cyclotron device is made to accelerate charge particles to extremely high speeds by applying crossed electric and magnetic fields.

<h3>Calculation of angular frequency:</h3>

Given,

B = 0.47 T

r = 0.68

mass of proton = 1.6x $10^{-27}$

q = 1.6 x $10^{-19}$

so, the frequency is:

f = qB/2 $\pi$ m

f = 1.6 x $10^{-19}$ x 0.47/2x3.14x1.6x $10^{-27}$

f = 0.07 x $10^{8}$

Hence, the angular frequency of the cyclotron is 0.07 x $10^{8}$ Hz.

Learn more about angular frequency here:

brainly.com/question/14244057

#SPJ4

7 0

2 years ago

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

3 years ago

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is </em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

3 years ago