(a) Let 
be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude
where 
is the radius of the circle.
The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is
so that
Solve for :
(b) The net force equation in part (a) leads us to the relation
so that is directly proportional to the square root of . As the radius increases, the maximum linear speed will also increase, so the cord is less likely to break if we keep up the same speed.
Internal energy of the system changes by ΔE = 178 J.
Heat given to the system = Q = +658 J.
According to the first law of thermodynamics,
ΔE = Q + W
178 = 658 + W
∴ W = 178-658 = -480 J
Minus sign indicates that work is done by the system.
Answer:
181.48 N
Explanation:
Calculate the area :
Area = pi * r² ;
pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m
Area 1, A1 = 3.14 * 0.1² = 0.0314 m²
Area 2, A2 = 3.14 * 0.9² = 2.5434 m²
Force, F = mass * acceleration due to gravity
F2 = 1500 * 9.8 = 14700 N
Force 1 / Area 1 = Force 2 / Area 2
Force 1 = (Force 2 / Area 2), * Area 1
Force 1 = (14700 / 2.5434) * 0.0314
Force = 5779.6650 * 0.0314
= 181.48 N
Answer:
20 m
Explanation:
Given:
v₀ = 15 m/s
v = -25 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(-25 m/s)² = (15 m/s)² + 2 (-10 m/s²) Δy
Δy = 20 m
<h3>16.</h3>
Your answer is correct.
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<h3>17.</h3>
The fractional change in resistance is equal to the given temperature coefficient multiplied by the change in temperature.
R = R₀×(1 + α×ΔT)
R = (10.0 Ω)×(1 + 0.004×(65 -20)) = 11.8 Ω
