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astra-53 [7]
3 years ago
15

A water balloon dropped from a window high above the ground falls y = 4.9t2 m in t sec. find the balloon’s (a) average speed dur

ing the first 3 sec of fall. (b) speed at the instant t = 3.
Physics
1 answer:
satela [25.4K]3 years ago
4 0
For the average speed, I multiplied 4.9 by 3 squared to receive 44.1, then divided that by 3 to receive 14.7 m/s. 4.9 X 4.9 X 4.9 = 44.1(Distance traveled / fall time)44.1 ÷ 3 = 14.7 m/s

For the speed at t = 3, I took the derivative of 4.9t2 to make it 2(4.9)t and then plugged in 3 to receive an answer of 29.4 m/s. derivative   =2(4.9)t                    =2(4.9)3                    =29.4 m/s
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In this problem, you will answer several questions that will help you better understand the moment of inertia, its properties, a
scoundrel [369]

Answer:

a)  Total mass form, density and axis of rotation location are  True

b)   I = m r²

Explanation:

a) The moment of inertia is the inertia of the rotational movement is defined as

       I = ∫ r² dm

Where r is the distance from the pivot point and m the difference in body mass

In general, mass is expressed through density

        ρ = m / V

        dm = ρ dV

From these two equations we can see that the moment of inertia depends on mass, density and distance

Let's examine the statements, the moment of inertia depends on

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- Acceleration angular False

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- axis of rotation location   True

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For a particle the mass is at a point whereby the integral is immediate, where the moment of inertia is

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4 0
3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



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