The displacement of the train after 2.23 seconds is 25.4 m.
<h3>Resultant velocity of the train</h3>
The resultant velocity of the train is calculated as follows;
R² = vi² + vf² - 2vivf cos(θ)
where;
- θ is the angle between the velocity = (90 - 51) + 37 = 76⁰
R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)
R² = 129.75
R = √129.75
R = 11.39 m/s
<h3>Displacement of the train</h3>
The displacement of the train is the change in position of the train after a given period of time.
The displacement is calculated as follows;
Δx = vt
Δx = 11.39 m/s x 2.23 s
Δx = 25.4 m
Thus, the displacement of the train after 2.23 seconds is 25.4 m.
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Answer:
Explanation:
The magnitude of the electrostatic force between two charged objects is
where
k is the Coulomb's constant
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:
is the distance between the charges
since the charges are identical
is the force between the charges
Re-arranging the equation and solving for q, we find the charge on each drop:
Answer:
d) 289.31 m
Explanation:
Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m
Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .
Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d
To equate
357.18 m -144.54 m = .735 m d
d = 289.31 m .
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