(a) The work done by the applied force is 26.65 J.
(b) The work done by the normal force exerted by the table is 0.
(c) The work done by the force of gravity is 0.
(d) The work done by the net force on the block is 26.65 J.
<h3>
Work done by the applied force</h3>
W = Fdcosθ
W = 14 x 2.1 x cos25
W = 26.65 J
<h3>
Work done by the normal force</h3>
W = Fₙd
W = mg cosθ x d
W = (2.5 x 9.8) x cos(90) x 2.1
W = 0 J
<h3>Work done force of gravity</h3>
The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.
<h3> Work done by the net force on the block</h3>
∑W = 0 + 26.65 J = 26.65 J
Thus, the work done by the applied force is 26.65 J.
The work done by the normal force exerted by the table is 0.
The work done by the force of gravity is 0.
The work done by the net force on the block is 26.65 J.
Learn more about work done here: brainly.com/question/8119756
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In our Solar System, Jupiter is the largest planet we have. it has the surface area of 23.71 billion mi^2. it beats all the other planets in both mass and volume.
Answer:
Stars are at large distances that even light takes thousands of years to reach us from there.
for example, alpha centauri is 4300 light years away from earth and it is considered the nearest star to us, this means that light from there takes 4300 years to reach us and with a spaceships that can move with the speed of light is would take us 4300 years to get there which is imposible to live for that long.
hence, it is difficult to move between the stars.
Answer:
I got you.. i'm in middle school and had that same question.
Explanation:
Refer to the diagram shown below.
The vertical distance traveled is
s = 25 m
The initial vertical launch velocity is zero.
Therefore
s = (1/2)*g*t²
where g = 9.8 m/s²
t = the time of flight, s
That is,
0.5*9.8*t² = 25
t² = 25/4.9 = 5.102
t = 2.26 s
Answer: 2.26 s
Answer:

Explanation:
As given point p is equidistant from both the charges
It must be in the middle of both the charges
Assuming all 3 points lie on the same line
Electric Field due a charge q at a point ,distance r away

Where
- q is the charge
- r is the distance
-
is the permittivity of medium
Let electric field due to charge q be F1 and -q be F2
I is the distance of P from q and also from charge -q
⇒
F1
F2
⇒
F1+F2=