Answer:
= 4.38 × 10³⁴kgm²/s
Explanation:
Given that,
mass of moon m = 9.5 × 10²²kg
Orbital radius r = 4.28 × 10⁵km
Orbital period T = 28.9days
T = 28.9 × 24 × 60 × 60
= 2,496,960s
Angular momentum of the moon about the planet
L = mvr
L = mr²w
If the two blocks are moving to the right at constant velocity, A. the horizontal force that B exerts on A points to the left. B
1 answer:
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The box has 3 forces acting on it:
• its own weight (magnitude <em>w</em>, pointing downward)
• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)
• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)
Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have
• net parallel force:
∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>
• net perpendicular force:
∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
<em>n</em> = <em>w</em> cos(35°)
<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)
<em>n</em> ≈ 120 N
Solve for the mag. of friction:
<em>f</em> = <em>µ</em> <em>n</em>
<em>f</em> = 0.25 (120 N)
<em>f</em> ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>
<em>a</em> ≈ (54.3157 N) / (15 kg)
<em>a</em> ≈ 3.6 m/s²
Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:
<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>
<em>v</em>² = 2 (3.6 m/s²) (3 m)
<em>v</em> = √(21.7263 m²/s²)
<em>v</em> ≈ 4.7 m/s
Answer:
5 ohms
Explanation:
Given:
EMF of the ideal battery (E) = 60 V
Voltage across the terminals of the battery (V) = 40 V
Current across the terminals (I) = 4 A
Let the internal resistance be 'r'.
Now, we know that, the voltage drop in the battery is given as:
Therefore, the voltage across the terminals of the battery is given as:
Now, rewriting in terms of 'r', we get:
Plug in the given values and solve for 'r'. This gives,
Therefore, the internal resistance of the battery is 5 ohms.