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lesantik [10]
3 years ago
11

Whats a non-example for average speed

Physics
2 answers:
jonny [76]3 years ago
6 0

A vector quantity is defined by magnitude and direction. For example, we might say that a car has an average speed of 25 miles per hour. Its average velocity might be 25 miles per hour due east.

Gnesinka [82]3 years ago
6 0

Answer:

Average speed is defined as the amount of movement that a object does in a given lapse of time.

So if i say: the car moves with a speed of 10km/h

This means that the car travels around 10km in the lapse of an hour, and here you can see that the speed is a scalar.

Now, if i say; the car moves at 10km/h to south.

We also have a direction, so we have now a vector, where the module of this vector is equal to the speed of the car.

so a non-example of average speed is the velocity of something.

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What aspect of motion can you conclude is common among freely falling objects ?
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Answer:

Free-fall is defined as the movement where the only force acting on an object is the gravitational force.

By the second Newton's law, we have that:

F = m*a

Where F = Force, m = mass, a = acceleration.

We can write this as:

a = F/m

And the gravitational force can be written as:

F = (G*M/r^2)*m

Where G is the gravitational constant, M is the mass of the Earth in this case, and r is the distance between both objects (the center of the Earth and the free-falling object)

As the radius of the Earth is really big, the term inside the parentheses is almost constant in the region of interest, then we can write:

G*M/r^2 ≈ g

And the gravitational force is:

F = g*m

And by the second Newton's law we had:

a = F/m = (g*m)/m = g

a = g

Then the acceleration does not depend on the mass of the object.

Then the thing that is common among the free-falling objects is the vertical acceleration.

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A spaceship is moving at 0.74 c (0.74 times the speed of light) with respect to the earth. An observer on the spaceship measures
olya-2409 [2.1K]

Answer:

The time interval is 22.76 hr.

Explanation:

Given that,

Time = 30 hrs

Speed = 0.74c

We need to calculate the time interval measured in earth frame

Using formula of time dilation

\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-\dfrac{v^2}{c^2}}}

Where, \Delta t = time interval measured in earth frame

\Delta t_{0}=proper time interval

Put the value into the formula

\Delta t=\dfrac{30}{\sqrt{1-\dfrac{0.74c^2}{c^2}}}

\Delta t=\dfrac{30}{\sqrt{1-(0.74)^2}}

\Delta t=\dfrac{30}{0.6726}

\Delta t=44.60\ hr

We need to calculate the proper time interval when the velocity is 0.86c

Using formula of time dilation

44.60=\dfrac{\Delta t_{0}}{\sqrt{1-\dfrac{0.86c^2}{c^2}}}

\Delta t_{0}=44.60\times\sqrt{1-(0.86)^2}

\Delta t_{0}=22.76\ hr

Hence, The time interval is 22.76 hr.

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