Answer:
A ball is thrown at an initial height of 5 feet with an initial upward velocity at 29 ft/s. lets assume that balls height h (in feet) after t seconds is give by:
<u>h= 5 + 29t -16t^2</u>
Explanation:
h= 5 + 29t -16t^2
a time when the ball's height will be 17 ft
17 = 5 + 29t -16t2
0 = -17 + 5 + 29t -16t2
0 = -12 + 29t - 16t2
Using the quadratic equation:
t = (-29±√(292-(4*(-16)*(-12))))÷2(-16)
= (-29±√(841 - 768))÷(-32)
= (-29±√(73))÷(-32)
= (-29 + 8.544)÷(-32) or (-29 - 8.544)÷(-32)
= (-20.456)÷(-32) or -37.544÷(-32)
= 0.64 or 1.17
So, the ball is at a height of 17 ft twice: once on the way up after 0.64 seconds and once on the way back down after 1.17 seconds.
Answer:
4.16 L
Explanation:
Assuming constant temperature,
At the edge of Typhoon Odessa: P₁ = 1 atm = 1013.3 mbar,
V₁ = 4.0 L
At the center of Typhoon Odessa: P₂ = (1013.3 - 40.0) mbar = 973.3 mbar
V₂ = ? L
For a fixed amount of gas at constant temperature (Boyle's law) : P₁V₁ = P₂V₂
V₂ = V₁ × (P₁/P₂)
V₂= (4.0) × (1013.3/973.3)
V₂= 4.16 L
Answer:
(a) The electron will move towards the wire.
The direction of the magnetic fields created by the wire can be found via right-hand rule. If you point your thumb towards the direction of the current, and if you curl your fingers, the direction of your four fingers will give the direction of the magnetic field. In this case, magnetic field is around the wire, and into the page just above the wire, where the electron is located.
According to the above formula, the direction of the force the wire applies to the electron can be found by right-hand rule.
Since the electron has a negative charge, the direction of the force is towards the wire.
(b) The proton will veer to the right.
The direction of the magnetic field is the same as the previous part. The proton has a positive charge, and coming from above. The direction of its velocity is downwards. The magnetic field above the wire is pointed into the page. Using the right-hand rule, the magnetic force on the proton is directed to the right, with respect to us.
The impulse given to the ball is equal to the change in its momentum:
J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s
This is also equal to the product of the average force and the time interval ∆t :
J = F(ave) ∆t
so that if F(ave) = 200 N, then
∆t = J / F(ave) = (2.8 kg•m/s) / (200 N) = 0.014 s
