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3 years ago

In the graph, which region shows nonuniform positive acceleration?

Physics

2 answers:

cricket20 [7]3 years ago

3 0

Answer: A.AB

Explanation:

This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.

As we can see in the attached image, the graph can be divided in four segments:

OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.

AB: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the acceleration is nonuniform and positive.

BC: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.

CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.

From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:

Segment AB

lesya692 [45]3 years ago

3 0

Answer:

AB plato users ;)

Explanation:

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Mercury has an average disease to the sun of 0.39 AU. In two or more complete sentences, explain how to calculate the orbital pe

alukav5142 [94]

Answer: 88 Earth days

Explanation:

According to the Kepler Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit:

$T^{2}=a^{3}$ (1)

If we assume the orbit is circular and apply Newton's law of motion and the Universal Law of Gravity we have:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (2)

Where $M$ is the mass of the massive object and $G$ is the universal gravitation constant. If we assume $M$ constant and larger enough to consider $G$ really small, we can write a general form of this law:

$MT^{2}=a^{3}$ (3)

Where $T$ is in units of Earth years, $a$ is in AU (1 Astronomical Unit is the average distane between the Earth and the Sun) and $M$ is the mass of the central object in units of the mass of the Sun.

This means when we are making calculations with planets in our solar system $M=1$ .

Hnece, in the case of Mercury:

$(1)T^{2}=(0.39 AU)^{3}$ (4)

Isolating $T$ :

$T=\sqrt{(0.39 AU)^{3}}$ (5)

$T=0.243 Earth-years \frac{365 days}{1 Earth-year}=88.6 days \approx 88 days$ (6)

This means the period of Mercury is 88 days.

7 0

3 years ago

The percent by which the fundamental frequency changed if the tension is increased by 30 percent is ? a)-20.04% b)-40.12% c)-30%

nika2105 [10]

Answer:

Percentage increase in the fundamental frequency is

d)-14.02%

Explanation:

As we know that fundamental frequency of the wave in string is given as

$f_o = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

now it is given that tension is increased by 30%

so here we will have

$T' = T(1 + 0.30)$

$T' = 1.30T$

now new value of fundamental frequency is given as

$f_o' = \frac{1}{2L}\sqrt{\frac{1.30T}{\mu}}$

now we have

$f_o' = \sqrt{1.3}f_o$

so here percentage change in the fundamental frequency is given as

$change = \frac{f_o' - f_o}{f_o} \times 100$

% change = 14.02%

3 0

3 years ago

2. A common physics experiment involves lowering an open tube into a cylinder of water and moving the tube up and down to adjust

notsponge [240]

Answer:

Explanation:

This question pertains to resonance in air column. It is the case of closed air column in which fundamental note is formed at a length which is as follows

l = λ / 4 where l is length of tube and λ is wave length.

here l = .26 m

λ = .26 x 4 = 1.04 m

frequency of sound = 330 Hz

velocity of sound = frequency x wave length

= 330 x 1.04

= 343.2 m /s

b )

Next overtone will be produced at 3 times the length

so next length of air column = 3 x 26

= 78 cm

c )

If frequency of sound = 256 Hz

wavelength = velocity / frequency

= 343.2 / 256

= 1.34 m

= 134 cm

length of air column for resonance

= wavelength / 4

134/4

= 33.5 cm

7 0

3 years ago

Which one of the following is a correct statement......

IceJOKER [234]

Answer:

Just 3

Explanation:

I believe the other two are incorrect

5 0

3 years ago

If a wave in a long rope has a period of 1.0 seconds and a wavelength of 0.2 what's the speed

Vilka [71]

brainly.com/question/243965 here is the answer this website

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4 years ago