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Veseljchak [2.6K]
3 years ago
8

(02.09 MC) An object starts at rest. Its acceleration over 30 seconds is shown in the graph below: An acceleration versus time g

raph is shown with acceleration on the y axis from zero to seven meters per second per second and time on the x axis from zero to 30 second. The graph starts at the origin and goes to five meters per second in the first five seconds, then goes back to zero at time 10 seconds, then stays at zero until 20 seconds, then goes up to 6 meters per second until 30 seconds. Use the graph above to determine the change in speed of the object between 20 and 30 seconds? (2 points) Select one: a. 6 m/s b. 30 m/s c. 60 m/s d. 180 m/s
Physics
2 answers:
Irina-Kira [14]3 years ago
3 0

Answer:

c. 60 m/s

Explanation:

As you can see in the graph, or the description of the graph, the change in speed since the acceleration at 20 seconds was 0 will be all the acceleration that we added up after that second, since we accelerated at 6 m/s until 30 seconds, that means that the object remained 10 seconds accelerating at 6m/s, so 10seconds times 6 m/s that is 60 m/s in the change in speed between the 20th second and the 30th second.

Verizon [17]3 years ago
3 0

Answer:

c. 60 m/s

Explanation: It makes sense if you think about it!

It took me a while but than it made sense!! I hope this helps you out!

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B. Projectile on cliff (range)
dimulka [17.4K]

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

         cos 25 = v₀ₓ / v₀

         sin 25 = Iv_{oy} / v₀

         v₀ₓ = v₀ cos 25

         v_{oy} = v₀ sin 25

         v₀ₓ = 22 cos 25 = 19.94 m / s

         v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

         y = y₀ + v_{oy} t - ½ g t²

     

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

         0 = 21 + 0.0192 t - ½ 9.81 t²

         4.905 t² - 0.0192 t - 21 = 0

         t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

        t = \frac{ 0.003914\  \pm \sqrt{0.003914^2 + 4 \ 4.2813 }   }{2}

        t = \frac{0.003914 \ \pm 4.13828}{2}

        t₁ = 2.07 s

        t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

        t = 2.07 s

now we can look up the distance traveled

         x = v₀ₓ t

         x = 19.94  2.07

         x = 41.28 m

5 0
3 years ago
A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20m\s. The bullet is brought to rest in a
prohojiy [21]

Answer:

1) F = 24 N

2) Distance = 1 m

Explanation:

We are given;

Mass; m = 120 g = 0.12 kg

Initial velocity; u = 20 m/s

Final velocity; v = 0 m/s since it came to rest.

Time; t = 0.1 s

We can calculate acceleration from Newton's first equation of motion;

a = (v - u)/t

a = (0 - 20)/0.1

a = -200 m/s²

1) magnitude of the resistance will be;

F = ma

F = 0.12 × (-200)

F = -24 N

Since, we are dealing with the magnitude, we will take the absolute value. Thus, F = 24 N

2) To find the distance moved by the bullet, we know that;

Distance = Average speed × time

Thus;

Distance = ((v + u)/2) × t

Distance = ((0 + 20)/2) × 0.1

Distance = 1 m

4 0
2 years ago
Why is a spectrum of colors produced when white light passes through a prism?
Mkey [24]

Explanation: White light is all colors of light in one, so when white light passes through a prism, the light gets refracted and breaks apart into all of the colors on the visible light spectrum.

4 0
2 years ago
A laser beam with a frequency of 180 Hz forms an 8 m standing wave with 10 nodes.
DIA [1.3K]

Answer:33

Explanation:

F = frequency

N =  Node count

w = wave lenght

v = wave velocity

L = distance wave traveled

First find wave length of laser

w = (2/(N))*(L)

w = (2/(10))*(8)

w = 1.6

then using (w), find velocity

V =  (w)(F)

V = (1.6)*(108)

V = 288

Plug in V and the new frequency to solve for new node count

F = NV/2L

(600) = (N)*(288) / 2 * (8)

(N) = 33.33

there are 33 nodes

8 0
3 years ago
Because of the curvature of the earth, the maximum distance D that you can see from the top of a tall building of height h is es
mojhsa [17]

Answer:

 D = 9.9 10⁶ mi

Explanation:

In the exercise they give the expression for maximum viewing distance

       D = 2 r h + h²

Ask for this distance for a height of 1100 feet

Let's calculate

        D = 2 3960 1100 + 1100²

        D = 8.712 10⁶ + 1.21 10⁶

        D = 9.92 10⁶ mi

         D = 9.9 10⁶ mi

8 0
3 years ago
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