Answer:
Explanation:
The relationship between the refractive index and the critical angle is given as follows:
where,
η = refractive index = 1.67
θc = critical angle =?
Therefore,
1. Uranium-238 decays by giving off one particle, and the decay can be represented by the following nuclear equation: 92238U 902
1 answer:
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Answer:
v = 1.98 mph
Explanation:
Given that,
Speed to travel one mile is 100 mph
Speed to travel another mile is 1 mph
The formula used to find your average speed is given by :
Putting the values, we get :
v = 1.98 mph
So, yours average speed is 1.98 mph.
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
