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Mila [183]
2 years ago
10

A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n

ext 25s. What distance did it travel during the first 45s?
Physics
1 answer:
Bogdan [553]2 years ago
8 0

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The<u> trajectory</u> is <u>the path followed by the body, the distance it travelled</u> (is a scalar quantity).  

The displacement is <u>the distance in a straight line between the initial and final position</u> (is a vector quantity).  

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m).  Being its displacement 95 m.

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Answer:

<em>4.61 N</em>

<em></em>

Explanation:

masa = 0.5 kg

ángulo de inclinación = 20°

Peso normal de la masa = mg

donde m = masa

g = aceleración debido a la gravedad = 9.81 m/s^2

Peso normal = 0.5 x 9.81 = 4.905 N

Si la masa se mantiene en su lugar mediante una cuerda paralela al plano, y no hay fricción en la masa, entonces

La fuerza sobre la cuerda = peso normal x cos ∅

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La fuerza sobre la cuerda = 4.905 x cos 20°

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4 0
2 years ago
While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the
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Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

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Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

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2 years ago
a 2.0*10^3 kg car accelerates from rest under the actions of two forces. One is a forward force of 1140 N provided by traction b
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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

<span>KE = 1/2mv^2 = (1/2)(2000)(2^2) = 4000 J This must equal the net work acting on the car. W=Fd The net force is 1140-950= 190N. so, d=W/F = 4000/190 = 21.05 m</span>
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2 years ago
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At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
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At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
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     = 314.37/144 = 2.18 lb/in²

Answer: 2.2 psi

5 0
2 years ago
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