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sweet [91]
3 years ago
6

A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2

and find your answer to the nearest 0.001 s.
Physics
1 answer:
MArishka [77]3 years ago
3 0
From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
                = 2×9.81×0.43
                = 8.4366
            u = √8.4366
               =2.905 m/s
Hence the initial velocity is 2.905 m/s
 Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
              = 2.905/9.81
              = 0.296 seconds


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Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

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m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

W=0.022\ kg\times 32\ ft/s^2=0.704\ lb

Hence, this is the required solution.

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Which data set has the largest range?
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Answer:

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Explanation:

The magnitude of the electrostatic force between two charged objects is

F=k\frac{q_1 q_2}{r^2}

where

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q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

r=0.070 cm =7\cdot 10^{-4} m is the distance between the charges

q_1=q_2=q since the charges are identical

F=9.0\cdot 10^{-9}N is the force between the charges

Re-arranging the equation and solving for q, we find the charge on each drop:

F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C

8 0
3 years ago
A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1m/s^2 for 20s.
SSSSS [86.1K]

Answer:

A.) 1430 metres

B.) 80 seconds

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Given that the train accelerates from rest at 1.1m/s^2 for 20s. The initial velocity U will be:

U = acceleration × time

U = 1.1 × 20 = 22 m/s

It then proceeds at constant speed for 1100 m

Then, time t will be

Time = distance/ velocity

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Time = 50 s

before slowing down at 2.2m/s^2 until it stops at the station.

Deceleration = velocity/time

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t = 22/2.2

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Using area under the graph, the distance between the two stations will be :

(1/2 × 22 × 20) + 1100 + (1/2 × 22 × 10)

220 + 1100 + 110

1430 m

The time taken between the two stations will be

20 + 50 + 10 = 80 seconds

6 0
3 years ago
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