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Ne4ueva [31]
3 years ago
15

Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo

city / change in time,” what is an abject acceleration at 4s? Be sure to include units in your math work and in your final answer.

Physics
1 answer:
sergiy2304 [10]3 years ago
6 0

Answer:

a=9\ cm/s^2

Explanation:

<u>Average Acceleration </u>

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: <em>We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that. </em>

When t_2=4\ sec, the graph shows a value of v_2=36\ cm/s

When t_1=0\ sec, the object is at rest, v_1=0

We compute the average acceleration as

\displaystyle a=\frac{v_2-v_1}{t_2-t_1}

\displaystyle a=\frac{36\ cm/s-0\ cm/s}{4\ sec-0\ sec}

\displaystyle a=\frac{36\ cm/s}{4\ s}

\boxed{a=9\ cm/s^2}

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When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted.
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A) We want to find the work function of the potassium. Apply this equation:

E = 1243/λ - Φ

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Given values:

E = 2.93eV, λ = 240nm

Plug in and solve for Φ:

2.93 = 1243/240 - Φ

Φ = 2.25eV

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E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

λ = 552nm

C) We want to find the frequency associated with the threshold wavelength. Apply this equation:

c = fλ

c = speed of light in a vacuum, f = frequency, λ = wavelength

Given values:

c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

Plug in and solve for f:

3×10⁸ = f(5.52×10⁻⁷)

f = 5.43×10¹⁴Hz

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Which two particles in an atom are equal in number
Lemur [1.5K]
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6 0
3 years ago
The x-coordinates of two objects moving along the x-axis are given as a function of time (t). x1= (4m/s)t x2= -(161m) + (48m/s)t
Kitty [74]
The two displacement functions are
x₁ = 4t
x₂ = -161 + 48t - 4t²
where
x₁, x₂ are in meters
t is time, s

The distance between the two objects is
x = x₁ - x₂
   =  4t + 161 - 48t + 4t²
x = 4t² - 44t + 161

Write this equation in the standard form for a parabola.
x = 4[t² - 11t] + 161
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Ths is a parabola that faces up and has its vertex (lowest point) at (5, 40).
Therefore the closest approach of the two objects is 40 m.
The graph of x versus t confirms the result.

Answer: The distance of the closest approach is 40 m.

5 0
3 years ago
A ball rolls 60 centimeters along a sidewalk in 5 seconds what is the speed of the ball.
meriva

Answer:

12 cm/s

Explanation:

Quite simply, you are looking for  cm/s

  so   60 cm / 5s   = 12 cm/s

6 0
2 years ago
Read 2 more answers
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