A 150kg motorcycle starts from rest and accelerates at a constant rate along with a distance of 350m. The applied force is 250N
1 answer:
Answer:
205N
Explanation:
The net force (F) is the difference between the applied force() and the kinetic frictional force(
). i.e
F = -
-----------------(i)
Note that;
= μmg
Where;
μ = coefficient of kinetic friction
m = mass of the body
g = acceleration due to gravity = 10m/s²
Equation (i) then becomes;
F = - μmg -------------------(ii)
<em>Given from question;</em>
m = mass of motorcycle = 150kg
μ = 0.03
= 250N
Substitute these values into equation (ii) as follows;
F = 250 - (0.03 x 150 x 10)
F = 250 - (45)
F = 205N
Therefore, the net force applied to the motorcycle is 205N
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Explanation:
Below is an attachment containing the solution.
Answer:
<em>Choice: c. 6sec</em>
Explanation:
<u>Horizontal Launch </u>
When an object is thrown horizontally with a speed (v) from a height (h), it describes a curved path ruled by gravity until it finally hits the ground.
The horizontal component of the velocity is always constant because no acceleration exists in that direction, thus:
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
Where
To calculate the time the object takes to hit the ground, we use the same formula as for free-fall, since the time does not depend on the initial speed:
The marble rolls the edge of the table at a height of h=180 m, thus:
t = 6 sec
Choice: c. 6sec
Answer:
(a) Charge of 4.25 μF capacitor is 35.46 μC.
(b) Charge of 1.13 μF capacitor is 10.05 μC.
(c) Charge of 2.85 μF capacitor is 25.36 μC.
Explanation:
Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:
C₄ = C₁ + C₂
Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.
C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF
Since, C₄ and C₃ are connected in series, there equivalent capacitance is:
C₅ =
Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.
C₅ =
C₅ = 2.05 μF
The charge on the equivalent capacitance is determine by the relation :
Q = C₅ V
Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.
Q = 2.05 μF x 17.3 = 35.46 μC
Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.
Charge on the capacitor, C₃ = 35.46 μC
Charge on the capacitor, C₄ = 35.46 μC
Voltage on the capacitor C₄ = =
= 8.90 volts
Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.
Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC
Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC