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ExtremeBDS [4]
4 years ago
15

What is the velocity of a car that travelled a total of 75 km north in 1.5 hrs?

Physics
1 answer:
AURORKA [14]4 years ago
3 0
V=d*t
Since d=75km=75000m
t=1.5h=3600*1.5= 5400s
=>V=75000*5400
=405000000m/s=4.05*10^8m/s
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Action reaction forces are in which one of newtons laws
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the third law

Explanation:

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A sailor pulls a crate across the deck of a ship with a rope, exerting a horizontal force of 150. N. The crate, which has a mass
nata0808 [166]

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3 years ago
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh
Dmitry_Shevchenko [17]

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = \frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

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It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

3 0
3 years ago
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