What causes diffferent air density?

A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph

Marina86 [1]

Solution:

initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m

h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º

8 0

3 years ago

The following are possible ways to express the quantity 0.391 (Give ALL correct answers, i.e., B, AC, BCD...) Note: 3.45E-8 is a

laiz [17]

To answer, evaluate the power of 10 in the given choices. If it is positve, move the decimal n places to the right. If it is negative, move the decimal n corresponding places to the left. From all the choices given, only the choices D, E, and F will give us the correct answer.

4 0

3 years ago

A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of

grigory [225]

Answer:

$a=2.9\ m/sec^2$

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force $F_r$ acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

$\vec F_r=m\vec a$

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force $F_r$ is computed as the hypotenuse of a right triangle

$|F_r|=\sqrt{256^2+104^2}$

$|F_r|=276.32\ Nw$

The acceleration can be obtained from the formula

$F_r=ma$

Note we are using only magnitudes here

$\displaystyle a=\frac{F_r}{m}$

$\displaystyle a=\frac{276.32Nw}{95.3Kg}$

$\boxed{a=2.9\ m/sec^2}$

7 0

3 years ago

An instrument that measures wind speed is called a(n ________.

Oduvanchick [21]

The instrument that measures wind is called anemometer. Anemometer not only measures wind, but also, this measures the direction of the wind. And anemometer is also a common device used in a station weather. It is dervived from the greek work anemos which means wind.

7 0

3 years ago

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

$U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m$

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J$

So, the required work done is 16 J.

7 0

3 years ago