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ycow [4]
3 years ago
9

You don't learn any movement concepts until high school.

Physics
2 answers:
stepan [7]3 years ago
5 0

Answer:

yes

Explanation:

you don't learn it until high school

hammer [34]3 years ago
4 0

Answer:

true

Explanation:

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A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s , releases a sandbag at an instant wh
lutik1710 [3]
<span>Data:

Initial velocity upward: Vo = 5.00 m/s ,
Initial position: h = 40.0 m above the ground

Type of motion: free fall.

A) Compute the position of the sandbag at a time 1.05 s after its release.

Equation: y = h + Vo*t - g*(t^2) / 2

y = 40.0 m + 5.00 m/s * 1.05s - (9.8 m/s^2) * (1.05 s)^2 / 2 = 39.8 m

B)Compute the velocity of the sandbag at a time 1.05 s after its release.

Equation: Vf = Vo - g*t

=> Vf = 5.00 m/s - (9.8m/s^2) * (1.05 s) = - 5.29 m/s


Negative sign means that the sandbag is going down.

c) How many seconds after its release will the bag strike the ground?

Equation:

y = yo + Vo*t - g*(t^2) / 2

0 = 40.0 + 5.00t - 4.9 t^2

=> 4.9 t^2 - 5t - 40 = 0

Use the quadratic formula and you get: t = 3.41 s
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3 years ago
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Can you build up a resistance to motion sickness
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Not that I know of. Sometimes watching something, listening to music, reading a book, or talking can help. There are some homeopathic remedies that can help as well.
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A forward horizontal force of 50 N is applied to crate a second horizontal force of 180 N is applied to crate in the opposite di
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As more and more bulbs are connected in series to a flashlight battery, what happens to the brightness of each bulb? Assuming th
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Answer:

P_1 = P_2 = \frac{P}{2}

so each bulb brightness becomes half of its given or indicated power

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Let the indicated power on the bulbs is given as P and its rated voltage is V

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R = \frac{V^2}{P}

now if the two bulbs are connected in series so we will have

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R_{eq} = 2\frac{V^2}{P}

now the current in the circuit is given as

i = \frac{V}{R_{eq}

i = \frac{P}{2V}

now brightness of each bulb is given as

P_1 = P_2 = i^2 R

P_1 = P_2 = \frac{P}{2}

so each bulb brightness becomes half of its given or indicated power

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P_1 = P_2 = \frac{V^2}{R}

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