Can at transform plate boundaries, two plate move toward each other true or false

a block of wood has a length of 4 cm a width of 5 cm and a height of 10 cm what is the volume of the wood

geniusboy [140]

Volume= Length X width X height.

Plug in the values for each and solve for the volume.

V= (L)(W)(H)

V=(4cm)(5cm)(10cm).

8 0

3 years ago

The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,

KATRIN_1 [288]

Answer:

The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.

C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'

Explanation:

See attached picture.

3 0

3 years ago

Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on

Zolol [24]

Answer:71 dB

Explanation:

Given

sound Level $\beta _1=124 dB$

distance $r_1=5.01 m$

From sound Intensity

$\beta =10dB\log (\frac{I_1}{I_0})$

$124=10dB\log (\frac{I_1}{I_0})$

$12.4=\log (\frac{I_1}{I_0})$

$I_1=(1\times 10^{-12})\times 10^{12.4}$

$I_1=2.51 W/m^2$

we know Intensity $I\propto ^\frac{1}{r^2}$

$I_1r_1^2=I_2r_2^2$

$I_2=I_1(\frac{r_1}{r_2})^2$

$I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2$

$I_2=1.24\times 10^{-5} W/m^2$

Sound level corresponding to $I_2$

$\beta _2=10\log (\frac{I_2}{I_0})$

$\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})$

$\beta _2=70.93\approx 71 dB$

6 0

3 years ago

at the center of the neutral metal block, what is the direction of the electric field contributed by the charge in and on the me

elena-14-01-66 [18.8K]

The direction of electric field by the charge in and on the metal block will be along the direction line 5 as given in question.

<h3>How to determine electric field direction in a metal block?</h3>

The charge always remain on outer surface of metal and inside the metal block, the net electric field is zero. But due to dipole there is an electric field at the center of metal block i.e. at point R along direction line 1.

Now, to make make the net electric field zero at center, the electric field by the charge in and on the metal block must be equal in magnitude to that of electric field due to dipole at point R and in opposite direction to that of the net electric field at at R due to dipole.

The electric field by the charge in and on the metal block will be making 180° angle to the electric field due to dipole at point R.

Hence the direction of electric field by the charge in and on the metal block will be along the direction line 5 as given in question.

To know more about electric field, click on brainly.com/question/11509296

#SPJ4

4 0

11 months ago

A disc of mass m slides with negligible friction along a flat surface with a velocity v. The disc strikes a wall head-on and bou

qwelly [4]

Answer:

-v/2

Explanation:

Given that:

a disc of mass m

Collides with the wall going through a sliding motion on on the plane smooth surface.

Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

<u>We know, kinetic energy is given as:</u>

$KE_i=\frac{1}{2}. m.v^2$

consider this to be the initial kinetic energy of the body.

<u>Now after collision:</u>

$KE_f=\frac{1}{4}\times KE_i$

$KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2$

Considering that the mass of the body remains constant before and after collision.

$KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2$

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.

3 0

3 years ago