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Answer:
2000 J
Explanation
Work equation is expressed as:

Where:
F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)
By substituting:

Have a nice day!
Answer:
move
Explanation:
i think that is it
as plates move large blocks of crust move along the faults
Answer:
The knights collide 53.0 m from the starting point of sir George.
Explanation:
The equation for the position in a straight accelerated movement is as follows:
x = x0 + v0 t + 1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial speed
a = acceleration
t = time
The position of the two knights is the same when they collide. Since they start from rest, v0 = 0:
Sir George´s position:
xGeorge = 0 m + 0 m + 1/2 * 0.300 m/s² * t²
Considering the center of the reference system as Sir George´s initial position, the initial position of sir Alfred will be 88.0 m. The acceleration of sir Afred will be negative because he rides in opposite direction to sir George:
xAlfred = 88.0 m + 0 m - 1/2 * 0.200 m/s² * t²
When the knights collide:
xGeorge = x Alfred
1/2 * 0.300 m/s² * t² = 88.0 m - 1/2 * 0.200 m/s² * t²
0.150 m/s² * t² = 88.0 m - 0.100 m/s² * t²
0.150 m/s² * t² + 0.100 m/s² * t² = 88.0 m
0.250 m/s² * t² = 88.0 m
t² = 88.0 m / 0.250 m/s²
t = 18.8 s
At t = 18.8 s the position of sir George will be
x = 1/2 * 0.300 m/s² * (18.8 s)² = <u>53.0 m </u>
Answer:
Explanation:
Given that, .
Pressure around scuba is
P = 10^5 Pa
1 Pa = 1 N/m²
Then
P = 10^5 N/m²
Descending height
h = 10m
Change in force per unit square centimetre
We know that,
Pressure = Force / Area
Then,
Force / Area is the required question we are finding
Then,
Force / Area = 10^5 N / m²
So, let convert the m² to cm²
100cm = 1m
(100cm)² = (1m)²
10⁴cm² = 1m²
Then,
Force / Area = 10^5 N/m² × 1m² / 10⁴cm²
Force / Area = 10 N/cm²
So, the force per unit square centimeters is 10.
In order to operate the theremin, a conducting object must be moved within the electric fields produced by the instrument