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Tpy6a
7 days 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:
Yuliya22 [2.4K]7 days ago
3 0

Answer:

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

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

Explanation:

A roller coaster features an extended ascent that enables the car to gain gravitational potential energy, which converts into kinetic energy, leading to a curve experience. In these curves, there are two sections: the lower section where riders experience heightened weight and the upper section where they feel lighter or weightless.

These feelings can be analyzed through Newton's second law; we will apply it to the lower section of the curve

         N - W = m a

the acceleration here is centripetal

        a = v² / r

substituting these values

        N = mg + m v² / r

        N = M (g + v² / r)

At this point, the perceived weight escalates due to the square of the body's speed, resulting in a significant feeling of heaviness.

In the top section of the curve, gravitational force still acts downward, and the normal force, which reacts to the surface, also pushes downwards, while the centripetal acceleration heads toward the curve's center, also directed vertically downward

        -N - W = -m a

         N = ma - W

         N = m (v² / r - g)

In this instance, the normal force, which generates the weight sensation, diminishes, leading to the perception of decreased weight. When v² / r equals g, the feeling of weightlessness becomes complete.

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You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Keith_Richards [2256]

Answer:

v₀ = 3.8 m/s

Explanation:

According to Newton's second law relating to the box:

∑F = m*a Formula (1)

∑F: the net force in Newton (N)

m: mass expressed in kilograms (kg)

a: acceleration measured in meters per second squared (m/s²)

Information known:

m = 2.1 kg, the mass of the box

d = 5.4m, the length of the roof

θ = 20° is the angle between the roof and the horizontal

μk = 0.51, the coefficient of kinetic friction between the box and the roof

g = 9.8 m/s², gravitational acceleration

Forces influencing the box:

The x-axis is oriented parallel to the box's movement on the roof, and the y-axis is oriented perpendicularly.

W: Weight of the box: directed vertically

N: Normal force: perpendicular to the roof's angle

fk: Frictional force: parallel to the direction along the roof

Calculating the weight of the box:

W = m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y components of weight:

Wx= Wsin θ=(20.58)*sin(20)°=7.039 N

Wy= Wcos θ=(20.58)*cos(20)°= 19.34 N

Finding the Normal force:

∑Fy = m*ay ay = 0

N-Wy = 0

N=Wy = 19.34 N

Calculating the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We substitute into Formula (1) to determine the box's acceleration:

∑Fx = m*ax ax=a: acceleration of the box

Wx-fk = (2.1)*a

7.039 - 9.86 = (2.1)*a

-2.821 = (2.1)*a

a=(-2.821)/(2.1)

a = -1.34 m/s²

Considering the box's Kinematics:

Since the box undergoes uniformly accelerated motion, we use the following to find the final speed of the box:

vf² = v₀² + 2*a*d Formula (2)

Where:

d refers to displacement = 5.4 m

v₀ is the initial speed

vf represents the final speed = 0

a is the box's acceleration = -1.34 m/s²

Plugging in the values into Formula (2):

0² = v₀² + 2*(-1.34)*(5.4)

2*(1.34)*(5.4) = v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

7 0
29 days ago
On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan
ValentinkaMS [2425]

Answer:

The equivalent distance in kilometers is 4012 ×10^{-6} km.

Explanation:

It's known that 1 millimeter converts to 10^{-3} meters. Then, 1 meter converts to 10^{-3} kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

1 mm = 10^{-3} m

1 m = 10^{-3} km

Thus, 1 mm = 10^{-3}×10^{-3} km = 10^{-6} km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

4012 mm = 4012 ×10^{-6} km.

The distance therefore is 4012 ×10^{-6} km.

5 0
1 month ago
A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f
kicyunya [2264]

Answer:

\Delta T = 0.81 ^oC

Explanation:

According to the principle of energy conservation

all kinetic energy will change into thermal energy to increase its temperature

\frac{1}{2}mv^2 = ms\Delta T

Next, divide both sides by the object's mass

\frac{1}{2}v^2 = s\Delta T

the resulting temperature change is expressed as

\Delta T = \frac{v^2}{2s}

\Delta T = \frac{25^2}{2\times 387}

\Delta T = 0.81^oC

3 0
12 days ago
A friend climbs an apple tree and drops a 0.22-kg apple from rest to you, standing 3.5 m below. When you catch the apple, you br
Keith_Richards [2256]
(A) velocity = 2.8 m/s (B) Average force = 1.9536 Newtons.
3 0
6 days ago
Select the areas that would receive snowfall because of the lake effect.
Keith_Richards [2256]

Result:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The locations likely to receive snowfall due to lake effect include Grand Marais, Two Harbors, and Duluth. These areas are situated directly along the shores of Lake Superior, one of the world’s largest lakes. Its substantial water volume significantly influences the local climate, generating a lot of humidity in the air and considerable evaporation, both of which lead to cloud formation and, when temperatures drop adequately, result in significant snowfall rather than rain.

8 0
6 days ago
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