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3 years ago

A car accelerates from 21.0 m/s to 46.0 m/s in 2.5 seconds. What is the acceleration of the car.

Physics

1 answer:

vivado [14]3 years ago

6 0

Answer:

$10.0 m/s^2$

Explanation:

The acceleration of an object is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval it takes for the velocity to change from u to v

For the car in this problem,

u = 21.0 m/s

v = 46.0 m/s

t = 2.5 s

Substituting, we find the acceleration

$a=\frac{46.0-21.0}{2.5}=10.0 m/s^2$

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notka56 [123]

Answer:

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Explanation:

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∅ = hc/λmax

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Therefore,

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Energy of Photon = ∅ + K.E

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Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

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Energy of Photon = hc/λ

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λ = wavelength of light = ?

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11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

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A car slows from 27 m/s to 5 m/s with a constant acceleration for 6.87 s. What is the car’s acceleration?

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In this case, the movement is uniformly delayed (the final rapidity is less than the initial rapidity), therefore, the value of the acceleration will be negative.

1. The following equation is used:

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Vf: final rapidity (m/s)

Vo: initial rapidity (m/s)

t: time (s)

2. Substituting the values in the equation:

a = (5 m/s- 27 m/s)/6.87 s

3. The car's acceleration is:

a= -3.20 m/ s<span>^2</span>

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sergij07 [2.7K]

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The magnitude of displacements a and b are 3m and 4m, respectively, c=a+b. What is the magnitude of c if the angel between a and

AysviL [449]

Answer:

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Explanation:

Given:

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(b)

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