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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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A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb
densk [106]
  • Voltage=V=120V
  • Power=P=60W

  • Current=I

\\ \rm\rightarrowtail I=\dfrac{P}{V}=\dfrac{60}{120}=\dfrac{1}{2}=0.5A

8 0
2 years ago
What is the velocity of a wave with a frequency of 45 Hertz and a wavelength of 3 meters?
Juli2301 [7.4K]

Explanation:

By using v=( f )x( lambda )

v= 45 s^-1 x 3 m

Therefore v = 135 ms^-1

8 0
2 years ago
A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If
inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

          v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

          v² = 2 a₁ x

let's calculate

          v = \sqrt {2 \ 15.0 \ 645}

          v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

          v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

         0 = v² - 2 a₂ x₂

         x₂ = v² / 2a₂

let's calculate

         x₂ = \frac{ 143.666^2 }{2 \  18.2}

         x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0
2 years ago
20. A cannon shoots a ball at 38 m/s at an angle of 30°. Ignore the original height of the cannon and assume level ground.
sergij07 [2.7K]
Answer:
a.3.87s
b.127.36m
c.18.4m

Step by step explanation:
Refer to the diagram

5 0
2 years ago
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