Answer is: <span>1/4 its old kinetic energy .
</span>V₁ = 10 m/s.
V₂ = 5 m/s.
m₁ = m₂ = m.
E₁ = 1/2 · m₁ · V₁², E₁ = 1/2 · m · (10 m/s)² = 50 · m.
E₂ = 1/2 · m₂ · V₂², E₂ = 1/2 · m · (5 m/s)² = 12,5 · m.
E₂/E₁ = 12,5m / 50m = 0,25.
V - speed of semi-truck.
m - mass of semi-truck.
E - kinetic energy of semi-truck.
Answer:
h = 48.077 ft
Explanation:
given,
distance between two observer = 300 ft
angle of elevation to top pole = 16° and 20°
height of the flagpole = ?
now,
Let h be the height of the flagpole
Let x be the distance of the pole
now,
again applying
h = 48.077 ft
Because no sunlight can penetrate it
Answer:
See the answer below
Explanation:
<em>The best thing one can do in this case would be to return the microscope's objective to low power and then </em><em>re-center the specimen </em><em>before switching back to high-dry power.</em>
Most of the time, <u>what makes the specimen under the microscope to be out of focus at higher objective powers after being in focus at low power is because they are not properly centered on the stage</u>. Hence, before calling on the instructor, it would be wise to first return to low power, re-center the specimen and bring it into focus after which the high power objective can be returned to and the fine focus adjusted to bring the image back to focus.
After doing the above and the specimen still does not come into focus, then the instructor can be called upon.
The LC circuit will also need a capacitance of 1.01 μF.
<h3>
What is capacitance?</h3>
Capacitance is the ratio of the amount of electric charge stored on a conductor to a difference in electric potential.
<h3>Capacitance needed for the resonance circuit</h3>
Xl = Xc
where;
- Xl is the inductive reactance
Xl = ωL
Xc = 1/ωC
ωL = 1/ωC
ω²LC = 1
C = 1/(ω²L)
C = 1/(2πf)²(L)
C = 1/(4π²f² L)
C = 1/(4π² x 100² x 2.5)
C = 1.01 x 10⁻⁶ F
C = 1.01 μF
Thus, the LC circuit will also need a capacitance of 1.01 μF.
Learn more about capacitance here: brainly.com/question/13578522
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