PLEASE HELP ):
1 answer:
The first thing we must do for this case is the sum of forces in a horizontal direction.
We have then:
Substituting values we have:
From here, we clear the mass of the object:
We now look for the weight of the object.
Where,
g: acceleration of gravity (9.8 m/s^2)
Substituting values:
Answer:
the weight of the object is:
option 4
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Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
Answer:
The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.
Explanation:
<u>Step 1:</u> calculate the charge on each electrode
Given;
Electric field strength = 2.0 X 10⁶ N/C
The distance between the electrode = 1mm = 1 X 10⁻³ m
Electric field strength (E) = Force (F)/Charge (q)
where;
E is the electric field strength = 2.0 X 10⁶ N/C
K is coulomb's constant = 8.99 X 10⁹ Nm²/C²
r is the distance between the electrodes = 1 X 10⁻³ m
q is the charge in each electrode = ?
= 0.2225 X 10⁻⁹ C
The charge on each electrode is 0.2225 X 10⁻⁹ C
<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.
1 electron contains 1.602 X 10⁻¹⁹ C
So, 0.2225 X 10⁻⁹ C will contain how many electrons ?
= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)
= 1.4 X 10⁹ electrons
Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.