<h2>
a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>
Explanation:
a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.
1300 m east = 1300 i
2400 m north = 2400 j
Drops the penny from a cliff 640 m high = -640 k
Displacement of penny = 1300 i + 2400 j - 640 k
b) Displacement of man for return trip = -1300 i - 2400 j

Magnitude of his displacement = 2729.47 m
Answer:
The distance is 55.636 billion miles, or 528.2 AU.
Explanation:
Since the distance from the Sun to Neptune is 2.7818 billion miles, the distance from the Sun to Planet Nine would be 20 times that, which is:

or 55.636 billion miles.
Since 1 astronomical unit (AU) is 93 million miles, that distance is also:

Answer:
<h2>
14.66secs</h2>
Explanation:
Given the formula for calculating the depth in metres expressed as
depth in meters = ½ (1500 m/sec × Echo travel time in seconds)
Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.
10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds
10,994 = 750 * Echo travel time in seconds
Dividing both sides by 750;
Echo travel time in seconds = 10,994 /750
Echo travel time in seconds ≈ 14.66secs (to two decimal places)
Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back
Answer:
410 m
Explanation:
Given:
v₀ = 20.5 m/s
a = 0 m/s²
t = 20 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (20.5 m/s) (20 s) + ½ (0 m/s²) (20 s)²
Δx = 410 m
1) Current in each bulb: 0.1 A
The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

And so, the current through the circuit is (using Ohm's law):

And since the two bulbs are connected in series, the current through each bulb is the same.
2) 4 W and 8 W
The power dissipated by each bulb is given by the formula:

where I is the current and R is the resistance.
For the first bulb:

For the second bulb:

3) 12 W
The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:
