You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba
1 answer:
Answer:
The ball reaches Barney head in
Explanation:
From the question we are told that
The rise velocity is
The height considered is
The horizontal velocity of the large object is
Generally from kinematic equation
Here s is the distance of the object from Barney head ,
u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter
So
So
=
Solving the above equation using quadratic formula
The value of t obtained is
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Answer: Graph C is the correct option
Explanation:
The question is incomplete, please remember to submit the whole question :)
However, the rest of the question with its corresponding figures is below:
Assume that positive momentum is to the right, which of the following best represents the momentum of the cart of mass m as a function of time before and after the collision?
The initial momentum of the cart with mass
(before the collision) is:
(1) Note the negative sign indicates the direction of cart's velocity (to the left, as seen in the first image attached)
On the other hand, the final momentum of both carts (after the inelastic collision) is:
(2)
So, according to this, the correct graph tha best represents the situation is C. Since before the collision the momentum is negative, then both carts slow down after the collision (), and taking into account the linear momentum is directly proportional to the velocity
(although is in the positive direction) is less than
.
Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance
∴ Area of each plate is 7.9060 m²
Voltage
∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb