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Tcecarenko [31]
2 years ago
14

For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir

st hill, where a horizontal spring that is initially compressed 0.25 m will push a small car forward. Each group of students will choose a car and a spring to push the car and then build a track. The assignment is to make the car go 5.0 m/s when it reaches the bottom of the first hill. Four groups of students choose springs and build tracks as described in the table. Which group’s roller coaster will most likely make the car travel closest to 5.0 m/s when it is at the bottom of the first hill?
Group Car Mass (Kg) Spring Constant (N/m) Hill Height (m)
A .75 kg 65 N/m 1.2 m
B .60 kg 35 N/m .9 m
C .55 kg 40 N/m 1.1 m
D .84 kg 32 N/m .95 m
Physics
2 answers:
abruzzese [7]2 years ago
7 0

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




Bas_tet [7]2 years ago
5 0

The answer is C. I just took the test on edge.

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You are standing at the midpoint between two speakers, a distance D away from each. The speakers are playing the exact same soun
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Answer:

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When he walks a distance of 1 m , path difference created = 2m

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Again we he walks a distance of 1.5 m , path difference created = 3 m

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In this way we see that man starts  from a point of maximum sound intensity , reaches a point of minimum sound intensity , then reaches a point of maximum sound intensity . At last he reaches a position of minimum sound intensity.

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2 years ago
A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.
Airida [17]

Answer:

Part A :  E =   \frac{1}{4\pi}ε₀ Q₁/R₁² Volt/meter

Part B :  V =  \frac{1}{4\pi}ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

<em>Part A</em>

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.  

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

                          F = \frac{1}{4\pi}ε₀ Q₁/R₁²

Then the electric field at that point is

                                   E =  F/1

                            E =  \frac{1}{4\pi}ε₀ Q₁/R₁²  Volt/meter

Part B

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

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4 0
1 year ago
A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the
lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

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c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

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0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

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Hope this Helps!!!

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