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Vlad [161]
3 years ago
9

Write a conclusion to this lab in which you discuss when a person on a roller coaster ride would have sensations of weightlessne

ss and when they would have sensations of weightiness. In your discussion, talk about accelerations and forces. Then finish off your conclusion by using Newton's second law to explain why such accelerations and force conditions cause these sensations.
Physics
1 answer:
masya89 [10]3 years ago
3 0

Answer:

he lower part of the curve     N = M (g + v² / r)

upper part of the curve          N = m (v² /r -g)

Explanation:

In a roller coaster there is a long climb that allows the car to acquire gravitational potential energy, when this energy is converted into kinetic energy, there is a raven, in these curves we have two parts the lower part, where you have a feeling of great weight and another in the upper part where you have a feeling of weightlessness.

These sensations can be explained using Newton's second law, let's apply it to the lower part of the curve

         N-W = m a

acceleration is centripetal

        a = v² / r

we substitute

        N = mg + m v² / r

        N = M (g + v² / r)

In this part the apparent weight is increased by the speed of the body squared, it feels like a lot of fart.

In the upper part of the curve the force of gravity continues to act downwards, the normal that is the reaction of the surface also goes downwards, the centripetal acceleration pointing towards the center of the curve has a vertical downward direction

        -N -W = -m a

         N = ma -W

         N = m (v² / r -g)

In this case we see that the normal that gives the sensation decreases, which is why we feel a loss of weight, in the case of v2 / r = g, the request is total and the sensation of weightlessness.

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A stone is thrown vertically upwards with a speed of 30.0 m/s.
matrenka [14]

a)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = final position of stone = 20.0 meters

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = ?

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 30² + 2 (- 9.8) (20 - 0)

v = 22.5 m/s


b)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = maximum height gained

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = 0 m/s

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

0² = 30² + 2 (- 9.8) (Y - 0)

Y = 46 m



6 0
3 years ago
PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!
pav-90 [236]

Answer:

50 N

Explanation:

Let the natural length of the spring = L

so

100 = k(40 - L)       (1)

200 = k(60 - L)       (2)

(2)/(1):   2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

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It says find the slope for each line I'm stuck on number one can you help me
Allushta [10]
\text{slope}=\frac{y_2-y_1}{x_2-x_1}

(-2, -1)(3, 1)

Therefore,

\text{slope}=\frac{1-(-1)}{3-(-2)}=\frac{1+1}{3+2}=\frac{2}{5}

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suppose you have a 69.0-kg wooden crate resting on a wood floor. what maximum force can you exert horizontally on the crate with
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the acceleration brought on by the gravitational pull of large masses generally, gravitational , often known as the acceleration brought on by the Earth's gravitational pull and centrifugal force,

F= friction coefficient *M*g

F= 0.5*69*9.8

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Likurg_2 [28]

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