Answer:
a. 12.12°
b. 412.04 N
Explanation:
Along vertical axis, the equation can be written as
T_1 sin14 + T_2sinA = mg
T_2sinA = mg - T_1sin12.5 ....................... (a)
Along horizontal axis, the equation can be written as
T_2×cosA = T_1×cos12.5 ......................... (b)
(a)/(b) given us
Tan A = (mg - T_1sin12.5) / T_1 cos12.5
= (176 - 413sin12.5) / 413×cos12.5
A = 12.12 °
(b) T2 cosA = T1 cos12.5
T2 = 413cos12.5/cos12.12
= 412.04 N
Answer:
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Answer:
potential energy PE = M g h
KE at bottom = 1/2 M V^2
Regardless of the slope of the slide the change in energy is the same
1/2 V^2 = g h
V = (2 g h)^1/2 = (2 * 9.8 m/s^2 * 10 m)^1/2 = 14 m / s
Perhaps the question says that h = 55 * .1 = 5.5 m
Then V = (2 * 9.8 * 5.5) = 10.4 m/s
Answer:
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Explanation:
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