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CaHeK987 [17]
3 years ago
11

When does an object falling vertically through the air reach terminal velocity?

Physics
1 answer:
zlopas [31]3 years ago
7 0

Answer:

a. when the acceleration of the objects become negative

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A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When th
aev [14]

Answer:

x_2=1.60m

Explanation:

From the Question We are told that

Initial Force F_1=5800N

Final Force F_2=6500N

Distance between the front and rear wheels \triangle x=3.20 m

Since

 \triangle x=3.20 m

Therefore

 x_1+x_2=3.20

 x_1=3.20-x_2

Generally the equation for The center of mass is at x_2 is mathematically

given by

 x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}

 x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}

 2*F_1*x_2 =3.20F_1

 x_2=1.60m

6 0
3 years ago
The gravitational attraction between two objects increases if
JulijaS [17]

Answer:

they are closer!!

Explanation:

Hope this helped!! :D

8 0
3 years ago
In Paragraph 3 of this passage, what clues help you know the meaning of the word
n200080 [17]
The answer is d because you have to make sure that everything is right
4 0
4 years ago
The temperature at one of the Viking sites on Mars was found to vary daily from -90.OF to -5.0 C. Convert these temperatures to
kykrilka [37]

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I don't know about this, this is which class question

6 0
3 years ago
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
3 years ago
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