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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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Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

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In order to find the electric charge towards x direction

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Charge diagonal:

e = kq/r²

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X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

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Sophie throws a tennis ball down from a height of 1.5 m at an angle of 450 with respect to vertical. She drops another tennis ba
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Given that,

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