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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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One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
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Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

3 0
3 years ago
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