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riadik2000 [5.3K]
3 years ago
15

A train traveling at 48 m/s begins to slow down as it approaches a bend in the tracks. If it travels around the bend at a speed

of 14 m/s, and it takes 40 seconds to properly slow down, what is the acceleration acting on the train during this time?
Please help 5 minutes left please!,
Physics
1 answer:
PSYCHO15rus [73]3 years ago
3 0

Answer:

Acceleration is (48-14)/40 m/s² = 34/40 m/s² = 0.85 m/s²

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which of the following objects is considered negatively charged A - one with excess electrons on the surface B - one with excess
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An object that could be considered as negatively charged would be when it has an excess of an electron in its atom. However, when it loses an electron, it could go back to its stable state which is "uncharged" or when there is an excess proton, it could be a positively charged object.
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When the wind kicks up dust and sand, the dust grains are charged. The small grains tend to get a negative charge, and the large
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Answer:

Explanation:

Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.

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3 years ago
In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little
horrorfan [7]

Answer:

t_up / t_down = 6.83

Explanation:

Find:

Calculate the ratio of the time he is above y_max/2 to the time it takes him to go from the floor to that height.

Solution:

- Compute the velocity v_o at y_max:

                             v_i^2 = v_f^2 - 2*g*y_max

                             0 = v_o^2 - 2*g*y_max

                             v_o = sqrt (2*g*y_max)

- The total time spend by athlete above height y_max / 2 is:

                             y - y_o = v_o*t_up - 0.5*g*t^2_up

                             v_o = 0.5*g*t_up

- Equate two equations:

                             sqrt (2*g*y_max) = 0.5*g*t_up

                             t_up = 2*sqrt(2*g*y_max) / g

- The total time taken by athlete to reach height y_max / 2 from ground is:

                            y - y_o = v_o*t_down - 0.5*g*t^2_down

                            g*t_^2down - 2*v_o*t_down + y_max = 0

- Solve the quadratic and evaluate t_down:

                            t_down = (v_o +/- sqrt (v^2_o - g*y_max)) / g

Substitute for v_o =  sqrt (2*g*y_max)

                            t_down = (sqrt(2g*y_max) +/- sqrt(g*y_max)) / g

- We will use the minus quantity, because we need the first part of the journey from ground from the two times he passes the height of y_max/2.

Hence,

                             t_down = (sqrt(2g*y_max) - sqrt(g*y_max)) / g

                             t_down = (sqrt(g*y_max) / g) * (sqrt(2) - 1)

- Compute the ratio t_up to t_down:

         t_up / t_down = 2*sqrt(2*g*y_max) / g * g / (sqrt(g*y_max)*(sqrt(2) - 1)

                                  = 2*sqrt(2) / (sqrt(2) - 1)

                                  = 6.83

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