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3 years ago

An object is thrown straight up into the air at 10 m/s. how fast is the object traveling after 2 seconds?

Physics

1 answer:

ziro4ka [17]3 years ago

7 0

First write down all the known variables:

vi = 10m/s

t = 1

a = -9.8m/s^2

vf = ?

Now choose the kinematic equation to relate these variables:

vf = vi + at

vf = (10) + (-9.8)(1)

vf = 0.2 m/s

The speed of the object travelling after a second would be 0.2m/s in reality, however, it is closest to the choice A, which would suffice as the answer for your question.

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Answer:

Yes it is possible because 3.33 is greater than 2.

Explanation:

Given that,

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Suppose water is leaking from the tank at a rate that is greater than 2 milliliters per second.

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2 years ago

A) Julie is running to the right at 5 m/s, as shown in the 1st figure. Balls 1 and 2 are thrown toward her at 10 m/s by friends

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Part 1)

here Julie is running at speed 5 m/s

So here two balls are thrown at speed 10 m/s towards Julie with respect to her friend standing on the ground.

So here this all speed is real speed of all.

Now as per Anita (let say she is one of her friend standing on ground) the speed of two balls will be same as the given speed as she is observing from ground or stationary frame

As per the frame of Julie

speed of ball 1

$v_{1j} = v_1 - v_j$

$v_{1j} = 10 - 5 = 5 m/s$ towards her in same side

speed of ball 2

$v_{2j} = v_2 - v_j$

$v_{2j} = -10 - 5 = -15 m/s$ towards her from opposite side

Part b)

Now in this case the speed of two balls is given with respect to Julie

so we can say

for ball 1

$v_{1j} = v_1 - v_j$

$10 = v_1 - 5$

so in ground frame speed of ball 1 is

$v_1 = 15 m/s$

Similarly for ball 2

$v_{2j} = v_2 - v_j$

$-10 = v_2 - 5$

So speed in ground frame of ball 2 is

$v_2 = - 5 m/s$

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3 years ago

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3 years ago

If stellar parallax can be measured to a precision of about 0.01 arcsec using telescopes on the Earth to observe stars, to what

marin [14]

Answer:

It corresponds to a distance of 100 parsecs away from Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

It is defined in a analytic way as it follows:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

$p('') = \frac{1}{d}$ (1)

Equation (1) can be rewritten in terms of d:

$d(pc) = \frac{1}{p('')}$ (2)

Equation (2) represents the distance in a unit known as parsec (pc).

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).

For the case of ( $p('') = 0.01$ ):

$d(pc) = \frac{1}{0.01}$

$d(pc) = 100$

Hence, it corresponds to a distance of 100 parsecs away from Earth.

<em>Summary:</em>

Notice how a small parallax angle means that the object is farther away.

Key terms:

Parsec: Parallax of arc second

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3 years ago