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8 days ago

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.

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The flight of a kicked football follows the quadratic function f(x)=−0.02x2+2.2x+2, where f(x) is the vertical distance in feet

Keith_Richards [3271]

The ball covers a horizontal distance of 0.902 meters. The trajectory of a kicked football adheres to a quadratic equation expressed as: f(x), where f(x) indicates the vertical distance in feet, and x signifies how far the ball travels horizontally. To compute the distance the ball will advance before striking the ground, we set the condition f(x) = 0. Upon solving this quadratic equation, we find that the horizontal distance traveled by the ball is: x = -0.902 meters, leading us to conclude that it travels 0.902 meters across the field.

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2 months ago

A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total

Ostrovityanka [3204]

Answer:

0.000047N

Explanation:

We know that

intensity (I) = P/ A

Where

P= power

A= Area

Thus, the power absorbed can be calculated as:

Power = Intensity x Area

This equals = 1.4 x 10^3 x(10)

Thus,

14000 Watts = 14 kWatt

However, the radiation pressure can be defined as

time-averaged intensity divided by the speed of light in a vacuum

So,

P = (1.4 x 1000)/c

Also,

F= P x A

Thus,

((1.4 x 1000)/(3 x10^8)) x 10

This results in

=0.000046666N

Rounded to two significant figures gives us

=0.000047 N

3 0

3 months ago

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0

serg [3582]

Since you've completed parts a and b, I will tackle part c.
For part C
To respond to this question, we must identify the zeros of the velocity function:
$v(t)=0.04t^3-0.06t^2$
This polynomial can be factored:
$v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)$
Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.
$t^2=0;\\ 0.04t-0.06=0$
By solving these equations, we identify our zeros:
$t_1=0; t_2=\frac{3}{2}$
The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:
$\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}$
From the table, it's evident that our function is positive when $- \infty < t$ and t>3/2.
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:
$s(t)=0.01t^4 - 0.02t^3$
We simply substitute t=12 to calculate the total distance traveled:
$s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft$
For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.
$a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t$
To find the acceleration at 1 second, we substitute t=1s into the previous equation:
$a(1)=0.12-0.12=0$

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2 months ago