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nataly862011 [7]
3 years ago
8

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the le

ft. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.
Physics
1 answer:
Nana76 [90]3 years ago
6 0

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

Hence angular velocity = 6/8=0.75rad/s

acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

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A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th
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\boxed{\text{\sf \Large 3.0 s}}

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It's photoelectric effect was fully explained by Einstein by the expression

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40/4.2=2*0.03*∆t

=158.73°C

7 0
3 years ago
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