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3 years ago

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According to the video, what tasks do Carpenters commonly perform? Check all that apply. supervising Plumbers, Electricians, and

Roofers forming walls and flooring out of concrete cutting, shaping, and fastening wood installing rafters, joists, windows, and subflooring operating heavy equipment setting hardwood floors building kitchen cabinets painting interior walls and cabinets

1 answer:

Marat540 [252]3 years ago

5 0

Answer:

Cutting, shaping, and fastening wood.

Installing rafters, joists, windows, and subflooring.

Setting hardwood floors.

Building kitchen cabinets.

Explanation:

Got the answer wrong so I could give you the right one :,)

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3. A penguin waddles 8 m uphill before sliding back down to its friends in 2 seconds. If the penguin ends where it started, what

german

The velocity of penguin as he ends where he started was 0 m/s.

<h3>What is displacement?</h3>

Displacement is the length of straight line joining the initial and final position of the body.

Given is a penguin who waddled 8 m uphill before sliding back down to its friends in 2 seconds.

We know that the velocity is the rate of change of displacement with respect to time. Mathematically -

v = dx/dt

dx = v dt

∫dx = ∫v dt

Δx = vΔt

v = Δx/Δt

Now, the displacement of the penguin will be = Δx = 8 - 8 = 0

Then, its velocity will be -

v = 0/Δt = 0

Therefore, the velocity of penguin as he ends where he started was 0 m/s.

To solve more questions on kinematics, visit the link below-

brainly.com/question/27200847

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1 year ago

A proton is projected into a magnetic field that is directed along the positive x axis. Find the direction of the magnetic force

bulgar [2K]

Te direction of the magnetic force for the velocity of the proton in the

-ve y direction will be +ve z direction.

As we know that the right-hand rule is based on the relation of magnetic fields and the forces that they exert on moving charges.When a charged particle moves under a magnetic field, it exerts a force on the particle, which is not in the same direction but different than the direction of the magnetic field.Under the right-hand rule, if we point our pointer finger in the direction of the charged particle is moving and the middle finger is representing the direction of the magnetic field then our thumb depicts the direction of the magnetic force which is exerted on the charged particle.

So, we are given that the direction of the velocity of the proton is in the negative y direction and the direction of the magnetic field is in the positive x direction, so the magnetic force is acting in the positive z direction.

To know more about the right-hand rule refer to the link brainly.com/question/9750730?referrer=searchResults.

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2 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago

How long does it take to change a flat plain into mountains? help

Len [333]

Im not for sure but i think it takes a couple hundred years (or according to the climate)

5 0

3 years ago

Read 2 more answers

What is the slope of the line plotted below?<br> A. 1.33<br> B. -1.33<br> C. -0.6<br> D. 0.6

Naya [18.7K]

Answer:

C. -0.6

Explanation:

Line is passing through the points ( - 3, 1) & (2, - 2)

Slope of line

$= \frac{ - 2 - 1}{2 - ( - 3)} = \frac{ - 3}{2 + 3} = \frac{ - 3}{5} = - 0.6 \\$

4 0

3 years ago