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just olya [345]

3 years ago

13

A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin

der spins so that the riders move with a linear speed v. At a certain point the floor of the cylinder lowers and the people are surprised they don't slide down. The friction between the wall and their clothes is μs = 0.62. What is the minimum speed, in meters per second, that the cylinder must make a person move at to ensure they will "stick" to the wall?

1 answer:

Leya [2.2K]3 years ago

7 0

Answer:

v = 13.19 m / s

Explanation:

This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

X axis

N = m a

Centripetal acceleration is

a = v² / r

Y Axis

fr -W = 0

fr = W

The force of friction is

fr = μ N

Let's calculate

μ (m v² / r) = mg

μ v² / r = g

v² = g r / μ

v = √ (g r /μ)

v = √ (9.8 11 / 0.62)

v = 13.19 m / s

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The energy of a light wave is calculated using the formula
E = hc/λ
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λ is the wavelength
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6 0

3 years ago

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jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

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Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

&

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$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

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A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that incre

Evgen [1.6K]

Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of $1.20 \ \Omega$

Answer:

The current is $I = 0.0007 41 \ A$

Explanation:

From the question we are told that

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The initial magnetic field at $t_o = 0 \ seconds$ is $B_i = 0.500 \ T$

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The resistance is $R = 1.20 \ \Omega$

Generally the induced emf is mathematically represented as

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$I = \frac{\epsilon}{R }$

=> $I = \frac{0.000889}{ 1.20 }$

=> $I = 0.0007 41 \ A$

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Answer:

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