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ExtremeBDS [4]
2 years ago
7

Discuss what it means to view a moving objectt from a frame of reference. Supply an example that illustrates your explanation.

Physics
1 answer:
tekilochka [14]2 years ago
5 0
<u><em>what is frame of reference:</em>
</u><span>it is set of fixed points which are inter related on the basis of rest and motion..,, or an explanation of what an observer is doing..
</span><u><em>examples:</em>
</u><span>1 ,, if you are in motion and observing something then frame of reference is motion... and if you are at rest then frame of motion is rest...
</span><span>2 frame of reference may be place like if a passenger is sitting in a car then he is in motion with respect to the ground but he is at rest with respect to the car in which he is sitting ...
</span><span>3 also if a passenger is in car and he is tossing a coin and you are on the ground  ,, then for passenger it is in vertical motion but for you it is in parabolic arc .. hence there are two different views for two persons for the same thing i.e coin... :)</span>
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1) A rock is dropped from a cliff with a height of 200 m. With what velocity will the rock hit the ground
Komok [63]

A rock is dropped from a 200 m high cliff. How long does it take to fall (a) the first 100 m and (b) the last 50 m?

The basic equation you want is:

s=at22

Solving for t:

t=2sa−−−√

We’ll assume a=9.8 , so 2a−−√=14.9−−−√≈0.4518

So, for (a)s=100 , so t=0.4518100−−−√=4.518

The total time is 0.4518200−−−√≈6.389

The time to fall 150 m is 0.4518150−−−√≈5.533

So the time to fall the last 50 m is 6.389 - 5.533 = 0.856 seconds

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2 years ago
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3 years ago
How are systems different from industries? Use an example to support your answer.
jeka94

Answer:

Explanation:

An industrial system consists of inputs, processes and outputs. The inputs are the raw materials, labor and costs of land,transport, power and other infrastructure. The processes include a wide range of activities that convert the raw material into finished products.

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2 years ago
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a body of radius R and mass m is rolling horizontally without slipping with speed v. it then rolls us a hill to a maximum height
ki77a [65]

Answer:

mR²/2

Explanation:

Here is the complete question

An object of radius′

R′  and mass ′

M′  is rolling horizontally without slipping with speed ′

V′

. It then rolls up the hill to a maximum height h = 3v²/4g. The moment of inertia of the object is (g= acceleration due to gravity)

Solution

Since it rolls without slipping, there is no friction. So, its initial mechanical energy at the horizontal surface equals its final mechanical energy at the top of the hill.

Since the object is rolling initially, and on horizontal ground, it initial energy is kinetic and made up of rotational and translational kinetic energy.

So, E = K + K'

E = 1/2mv² + 1/2Iω² where m = mass of object, v = speed of object, I = moment of inertia of object and ω = angular speed of object = v/r where v = speed of object and R = radius of object.

Also, the final mechanical energy of the object, E' is its potential energy at the top of the hill. So, E' = mgh.

Since E = E',

1/2mv² + 1/2Iω² = mgh

substituting the values of ω and h into the equation, we have

1/2mv² + 1/2Iω² = mgh

1/2mv² + 1/2I(v/R)²= mg(3v²/4g)

Expanding the brackets, we have

1/2mv² + 1/2Iv²/R²= 3mv²/4

Dividing through by v², we have

1/2m + I/2R²= 3m/4

Subtracting m/2 from both sides, we have

I/2R² = 3m/4 - m/2

Simplifying, we have

I/2R² = m/4

Multiplying through by 2R², we have

I = m/4 × 2R²

I = mR²/2

6 0
2 years ago
Marissaâs car accelerates uniformly at a rate of +2.60 m/s². How long does it take for Marissaâs car to accelerate from a speed
DIA [1.3K]

Answer:

The time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds.

Explanation:

Given that,

Acceleration of the car, a=+2.6\ m/s^2

Initial speed of the car, u = 24.6 m/s

Final speed of the car, v = 26.8 m/s

We need to find the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s. The acceleration of an object is given by :

t=\dfrac{v-u}{a}

t=\dfrac{(26.8-24.6)\ m/s}{2.6\ s}

t = 0.84 seconds

So, the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds. Hence, this is the required solution.                                    

4 0
3 years ago
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