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ankoles [38]
3 years ago
10

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 228 kg and moves

with speed v = 16.74 m/s. The loop-the-loop has a radius of R = 11.3 m.
1.What is the magnitude of the normal force on the care when it is at the bottom of the circle? (But as the car is accelerating upward.)
2.What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)?
3.What is the magnitude of the normal force on the car when it is at the top of the circle?
4.Compare the magnitude of the cars acceleration at each of the above locations:

A. abottom = aside = atop
B. abottom < aside < atop
C. abottom > aside > atop
5.What is the minimum speed of the car so that it stays in contact with the track at the top of the loop?
Physics
1 answer:
Mariulka [41]3 years ago
8 0

Answer:

A.

ride. You estimate thatch tofind that each loopp

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Answer:

C) Index of refraction

Explanation:

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The formula for Index of refraction is:

n = c ÷ v

where n represents Index of refraction,

c represents speed of light in a vacuum,

v represents speed of light in a given material.

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How do the units that are used to measure heat differ from the units that are used to measure temperature?
irakobra [83]
Heat is measured in joules and temperature is measured in F degrees or C degrees.
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4 years ago
Suppose the block is released from rest with the spring compressed 5.00 cm. The mass of the block is 1.70 kg and the force const
IRISSAK [1]

First, let's calculate the total mechanical energy when the block is at rest and the spring is compressed 5 cm:

\begin{gathered} ME=PE+KE\\ \\ ME=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ ME=\frac{955\cdot0.05^2}{2}+0\\ \\ ME=1.194\text{ J} \end{gathered}

Now, let's use this total energy to calculate the velocity when the spring is compressed by 2.5 cm:

\begin{gathered} ME=PE+KE\\ \\ 1.194=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ 2.388=955\cdot0.025^2+1.7v^2\\ \\ 1.7v^2=2.388-0.597\\ \\ 1.7v^2=1.791\\ \\ v^2=\frac{1.791}{1.7}\\ \\ v^2=1.0535\\ \\ v=1.026\text{ m/s} \end{gathered}

Therefore the speed is 1.026 m/s.

5 0
1 year ago
A magnetic field is directed perpendicular to the plane of a 0.15-m × 0.30-m rectangular coil consisting of 240 loops of wire. T
ValentinkaMS [17]

Answer:

7.344 s

Explanation:

A = 0.15 x 0.3 m^2 = 0.045 m^2

N = 240

e = - 2.5 v

B1 = 0.1 T

B2 = 1.8 T

ΔB = B2 - B1 = 1.8 - 0.1 = 1.7 T

Δt = ?

e = - dФ/dt

e = - N x A x ΔB/Δt

- 2.5 = - 240 x 0.045 x 1.7 / Δt

2.5 = 18.36 / Δt

Δt = 7.344 s

6 0
3 years ago
What is the maximum speed when the conditions are mass =450 kg, initial height= 30 m, and the roller coaster is initially at res
Zarrin [17]

Answer:

B. 24.2 m/s

Explanation:

Given;

mass of the roller coaster, m = 450 kg

height of the roller coaster, h = 30 m

The maximum potential energy of the roller coaster  due to its height is given by;

P.E_{max} = mgh\\\\PE_{max} = 450 *9.8*30\\\\PE_{max} = 132,300 \ J

P.E_{max} = K.E_{max} \ (law \ of \ conservation\ of \ energy)

K.E_{max} = \frac{1}{2}mv_{max}^2\\\\ v_{max}^2 = \frac{2K.E_{max}}{m}\\\\ v_{max}^2 = \frac{2*132300}{450}\\\\ v_{max}^2 =588\\\\v_{max} = \sqrt{588}\\\\  v_{max} = 24.2 \ m/s

Therefore, the maximum speed of the roller coaster is 24.2 m/s.

3 0
3 years ago
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