The vector product of vectors A⃗ and B⃗ has magnitude 12.0 m2 and is in the +z-direction.Vector A⃗ has magnitude 4.0 m and is in
1 answer:
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Answer:
Explanation:
The general equation for position of Simple harmonic motion is given as:
........(1)
where,
x = Position of the wave
A = Amplitude of the wave
ω = Angular velocity
t = time
In this case, the amplitude is just half the range,
thus,
(Given range = 3cm)
A = 1.5 cm
Now, The angular velocity is given as:
Where, T = time period of the wave =0.27s (given)
or
so, at time t = 55 s, the equation (1) becomes as:
on solving the above equation we get,
here the negative sign depicts the position in the opposite direction of +x
Answer:
The shell will land 10.18m away from the buoy.
Explanation:
In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).
Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.
In order to do so we can use the following equation:
now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:
and solve for t:
now we can substitute the values:
t=3.19s
Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.
So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:
We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:
So we can substitute the given values:
Which yields:
so now we can use the A and B points to find by how far the shell missed the buoy:
Distance=329.18m-319m=10.18m
So the shell missed the buoy by 10.18m.
Answer:
U = 1794.005 × 10⁶ J
Explanation:
Data provided;
Capacitance of the original capacitor, C = 1.27 F
Potential difference applied to the original capacitor, V = 59.9 kV
= 59.9 × 10³ V
Now,
The Potential energy (U) for the capacitor is calculated as:
Potential energy of the original capacitor, U = × C × V²
on substituting the respective values, we get
U = × 1.27 × ( 59.9 × 10³ )²
or
U = 1794.005 × 10⁶ J