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vagabundo [1.1K]
3 years ago
8

The motion of a car on a position time graph is represented with a horizontal tine What does this indicate about the car's motio

n?
Physics
1 answer:
GalinKa [24]3 years ago
8 0

Answer:

Position-Time graphs display the motion of a object by showing the changes of velocity with respect to time.

The motion of a car on a position-time graph that is represented with a horizontal line indicates that the car has stopped moving.

A straight line with a positive slope indicates that the car is moving at a constant velocity, and thus the slope is constant. On the other hand, a curve with a changing slope, shows that the velocity is changing.

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Long wavelength corresponds to having _________frequency
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Long wavelength corresponds to having lower frequency
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Please who can give me the definition of density in terms of Archimedes principle.​
VladimirAG [237]

\large\bold{\underline{\underline{Answer:-}}}

  • The weight of the fluid displaced is equal to the buoyant force on a submerged object. The mass divided by the volume thus determined gives a measure of the average density of the object.
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2 years ago
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After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48
masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

15min*\frac{1hr}{60min}=0.25hr

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

V=\frac{distance}{time}

the plane traveled a distance to the north of 48km so the velocity is:

V=\frac{48km}{0.25hr}

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

V_{x}=42km/hr*cos(-19^{o} )=39.71 i

and

V_{y}=42km/hr*sin(-19^{o} )=-13.67 j

So the velocity of the wind can be expressed as a vector as:

V_{wind}=(39.71i - 13.67j)km/hr

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

V_{plane x}=V_{plane/wind x}+V_{wind x}

V_{plane/wind x}=V_{plane x}-V_{wind x}

V_{plane/wind x}=(0-39.71 i)km/hr

V_{plane/wind x}= -39.71 i km/hr

and

V_{plane y}=V_{plane/wind y}+V_{wind y}

V_{plane/wind y}=V_{plane y}-V_{wind y}

V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr

V_{plane/wind x}= 205.67 j km/hr

So the velocity of the plane with respect to the wind can be rewritten as:

V_{plane/wind x}= (-39.71i + 205.67 j) km/hr

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

speed=\sqrt{(-39.71)^{2}+(205.67)^{2}  }

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0
3 years ago
Can someone please give me the (Answers) to this? ... please ...<br><br> I need help….
IceJOKER [234]

#1.

<em>Car </em>1<em> weighs </em>300 kilograms<em> and is moving right at </em>3 meters per second (m/s)

  • v1 (before) = 3 m/s

  • v2 (before) = -1 m/s

  • v1 (after) = 0.5 m/s

#2.

Law of conservation of momentum

momentum before collorion = momentim after collosion

MV + mv = MV' + mv'

1500x25+ 1000x5

37500 + 15000

6 0
2 years ago
Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 390 N while the ot
romanna [79]

Answer:

The crate's coefficient of kinetic friction on the floor is 0.23.

Explanation:

Given that,

Mass of the crate, m = 300 kg

One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.

The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :

F_1+F_2=\mu N\\\\\mu=\dfrac{F_1+F_2}{mg}\\\\\mu=\dfrac{390+320}{300\times 10}\\\\\mu=0.23

So, the crate's coefficient of kinetic friction on the floor is 0.23.

6 0
3 years ago
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