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Ludmilka [50]
2 years ago
6

marissas car accelerates uniformly at a rate of -2.60m/s^2. how long does it take for marissas car to accelerate from a velocity

of -28.2m/s to -42.3m/s?
Physics
1 answer:
zlopas [31]2 years ago
7 0

Answer:

1.02s

Explanation:

In this situation the following equation will be useful:



Where:

 is Marissa's car  final velocity

 is Marissa's car initial velocity

 is Marissa's car constant acceleration  (assuming this is the acceleration, since 1269 m/s^{2} does not make sense)

 is the time  it takes to accelerato from  to 

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Explanation:

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Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i
mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2}  =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}

\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad

w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}

The velocity at the contact point is equal to:

v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}

v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}

Matching both expressions:

1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s

b) The time during which the disks slip is:

t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s

a) The angular acceleration of each disk is

\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)

\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)

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2 years ago
Only one symbol must be included in every circuit diagram. What symbol is it?
Kazeer [188]

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A symbol for a battery.

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A battery, because every sinlge circuit AC or DC must have a source of energy, to supply it to every single device that is part of the circuit.

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An electron at Earth's surface experiences a gravitational force of magnitude F=(9.11×10−31 kg)⋅(9.8 m/s2). Part A How far away
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Answer:

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Explanation:

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The electric force between an electron and a proton is given by Coulomb's Law

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They ask us that W = Fe

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Let's calculate

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Let's look for the relationship of this distance with the harmonic distance

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We see that this distance is 10¹⁰ times the interatomic distance, so the gravitational attraction force is very small at atomic scale

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