A bar of soap has a mass of 53.65 grams and its dimensions are 4.6cm by
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Answer:
The work done required on the coin during the displacement is 21.75 w.
Explanation:
Given that,
A coin slides over a friction-less plane i.e friction force = 0.
The co-ordinate of the given point is (1.40 m, 7.20 m).
The position vector of the given point is represented by .
The displacement of the coin is
The force has magnitude 4.50 N and its makes an angle 128° with positive x axis.
Then x component of the force = 4.50 cos128°
The y component of the force = 4.50 sin128°
Then the position vector of the force is
We know that,
work done is a scalar product of force and displacement.
=(-2.77×1.40+ 3.56×7.20) w
=21.75 w
The work done required on the coin during the displacement is 21.75 w.
Answer:
The ball will drop 0.881 m by the time it reaches the catcher.
Explanation:
The position of the ball at time "t" is described by the position vector "r":
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
When the ball reaches the catcher, the position vector will be "r final" (see attached figure).
The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.
Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):
x = x0 + v0x · t
17.0 m = 0 m + 40.1 m/s · t
t = 17.0 m/ 40. 1 m/s = 0.424 s
With this time, we can calculate the y-component of the vector "r final", the drop of the ball:
y = y0 + v0y · t + 1/2 · g · t²
Initially, there is no vertical velocity, then, v0y = 0.
y = 1/2 · g · t²
y = -1/2 · 9.8 m/s² · (0.424 s)²
y = -0.881 m
The ball will drop 0.881 m by the time it reaches the catcher.
