PHYSICS COLLEGE

Why is wood a renewable resource?A) We use it over and over again.
B) It takes millions of years to replace.
C) Trees grow everywhere, so we always have some.
D) It can be replaced in a reasonable amount of time.

Answers

Answer 1
Answer:

Answer:

D) It can be replaced in a reasonable amount of time.

Explanation:

i took the test!!!!!1

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A sphere has a charge of −84.0 nC and a radius of 5.00 cm. What is the magnitude of its electric field 3.90 cm from its surface?

Answers

Answer:

E = -9.5* 10^4~N/C

Explanation:

Gauss' Law should be applied to find the E-field 3.9 cm from the surface of the sphere.

In order to apply Gauss' Law, an imaginary spherical shell (Gaussian surface) should be placed around the original sphere. The exact position of the shell must be 3.9 cm from the surface of the original sphere.

Gauss' Law states that

\int {\vec{E}d\vec{a}} = (Q_(enc))/(\epsilon_0)

Here, the integral in the left-hand side is equal to the area of the imaginary surface. After all, the reason behind choosing the imaginary surface a spherical shell is to avoid this integral. The enclosed charge in the right-hand side is equal to the charge of the sphere, -84.0 nC. The radius of the imaginary surface must be 5 + 3.9 = 8.9 cm.

So,

E4\pi r^2 = (-84* 10^(-9))/(8.8* 10^(-12))\nE4\pi (8.9 * 10^(-2))^2 = (-84* 10^(-9))/(8.8* 10^(-12))\n\nE = -9.5* 10^4~N/C

The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Determine the location ӯ of the center of mass G of the pendulum, then calculate the mass moment of inertia of the pendulum about z axis passing through G. b) Calculate the mass moment of inertia about z axis passing the rotation center O.

Answers

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

y=(0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m)/((\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m) \n\ny=0.88\ m

b)  the mass moment of inertia about z axis passing the rotation center O is:

I_G=(1)/(12)*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\n0.888)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+(1)/(12)*3(1.5)(1.5)^2+\n3(1.5)(0.888-0.75)^2\n\nI_G=13.4\ kgm^2

c) The mass moment of inertia about z axis passing the rotation center O is:

I_o=(1)/(12)*3(0.8)(0.8)^2+ (1)/(3)* 3(1.5)(1.5)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\n(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\n\nI_o=13.4\ kgm^2

Final answer:

To solve this problem, calculate the mass of each element of the pendulum, use that information to determine the center of mass, and then apply the parallel axis theorem to calculate the two moments of inertia.

Explanation:

To determine the center of mass and the mass moment of inertia of the pendulum, first we calculate the individual masses of the rods: AB and OC, and the plate. Each rod has a mass of 2 kg (given mass per unit length is 3kg/m and length of each rod is 1 m from the first reference paragraph).

The center of mass ӯ can be determined using the formula for center of mass, averaging distances to each mass element weighted by their individual masses. The mass moment of inertia, also known as the angular mass, for rotation about the z axis through G is determined using the parallel axis theorem, which states that the moment of inertia about an axis parallel to and a distance D away from an axis through the center of mass is the sum of the moment of inertia for rotation about the center of mass and the total mass of the body times D squared.

Finally, the moment of inertia about the z axis passing through the center of rotation O can be calculated again using the parallel axis theorem, with distance d being the distance between points G and O.

Learn more about Mass Moment of Inertia here:

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Define reflection of sound?​

Answers

The reflection of sound is the movement of sound waves bouncing off of a surface and back into another direction, hope this helped :)

A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter stick hangs a 5.202 kg concrete block. From the 75.0 cm mark on the meter stick hangs a 7.99 kg steel ball. At what mark on the meter stick must a 2.46 kg wooden block be attached so that the meter stick balances horizontally when lowered into fresh water? Assume the densities of concrete, steel, and wood are 2500.0 kg/m3, 8000.0 kg/m3, and 500 kg/m3 respectively. A. 57.6 cm
B. 57.4 cm
C. 57.2 cm
D. 57.8 cm

Answers

The wooden block must be attached at the 57.6 cm mark on the meter stick.

What is the value

We can solve this equation for the distance of the wooden block from the pivot point, which is the mark on the meter stick where it must be attached so that the meter stick balances horizontally when lowered into fresh water.

The density of aluminum is 2700 kg/m³ The volume of the meter stick is 0.01 m³ (length times width times thickness). The acceleration due to gravity is 9.81 m/s²

Substituting these values into the equation above, we get:

distance of wooden block from pivot point = -44.89 Nm / [(2700 kg/m³)(0.01 m³)(9.81 m/s²)]

= 0.576 m

Learn more about meter

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Answer:

Stop cheating in exam

Explanation:

Shame!!!!

I am sorry but I will have to refer you to the student conduct at UTA.

Some snakes have special sense organs that allow them to see the heat emitted from warm blooded animals what kind of an electromagnetic waves does the sense organs detect?A. Visible light waves
B. Ultraviolet light waves
C. Infrared Waves
D. Microwaves
​

Answers

The heat emitted from anything is carried in the form of infrared waves. (C)

A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the ground. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.

Answers

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
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