True or false - The vector for a negative acceleration points in the opposite direction when

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

so

$\theta=5.71^{o}$

8 0

2 years ago

If an electron is lost or gained, we no longer call it an atom. What is the name when this happens?

gtnhenbr [62]

It is called an ion (positive or negative ion)

3 0

3 years ago

A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring balance C and D. Balance C is

vagabundo [1.1K]

A uniform metal rod with of length 80cm and a mass of 3.2kg is supported horizontally by two vertical spring balances C and D. Balance C is 20cm from one end while D is 30cm from the other end would show the reading of 1.06 Kg and 2.13 kg respectively

<h3>What is gravity?</h3>

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another, The gravity varies according to the mass and size of the body for example the force of gravity on the moon is the 1/6th times of the force of gravity on the earth.

As given in the problem, A uniform metal rod of the length of 80cm and mass of 3.2kg is supported horizontally by two vertical springs balance C and D. Balance C is 20cm from one end while D is 30cm from the other end

The weight of the rod acting downward is from the center of the rod at 40 cm

Let us suppose the reading on the spring balance C and D are P and Q respectively

By using the equilibrium for the vertical force

Fv=0

P + C = 3.2

By using the equilibrium for the moment around the left corner

20×P+ 50×Q= 40 ×3.2

By solving for both P and Q from the above two equations we would get

P =1.06 and Q = 2.13

Thus, the reading on the spring balance C and D would be 1.06 Kg and 2.13 kg respectively

Learn more about gravity from here

brainly.com/question/4014727

#SPJ1

5 0

1 year ago

A rectangular metal tank with an open top is to hold 171.5 cubic feet of liquid. what are the dimensions of the tank that requir

Semmy [17]

To minimize the material usage we have to have the volume requested with the minimum surface area.
The volume is:
$171.5 =xyz$
And the surface is:
$S=xy+2xz+2yz$
From the first equation we get:
$z=\frac{171.5}{xy} ; k=171.5\\ z=\frac{k}{xy}\\$
I will use k instead of a number just for the conveince.
We plug this into the second equation and we get:
$S=xy+2k\frac{1}{x}+2k\frac{1}{y}$
To find the minimum of this function we have to find the zeros of its first derivative.
Sx will denote the first derivative with respect to x and Sy will denote the first derivative with respect to Sy.
$S_x=y-2k\frac{1}{x^2}\\ S_y=x-2k\frac{1}{y^2}$
Now let both derivatives go to zero and solve the system (this will give us the so-called critical points).
$0=y-2k\frac{1}{x^2}\\
0=x-2k\frac{1}{y^2}\\
y=2k\frac{1}{x^2}\\
x=2k\frac{1}{y^2}\\$
Now we plug in the first equation into the other and we get:
$x=\frac{\frac{2k}{1}}{\frac{4k^2}{x^4}}\\
x^3=2k\\
x=(2\cdot171.5)^{1/3}\\
x=7
$
Now we can calculate y:
$y=2k\frac{1}{x^2}\\
y=2\cdot 171.5\frac{1}{7^2}=7$
And finaly we calculate z:
$z=\frac{171.5}{xy}\\
z=\frac{171.5}{7\cdot7}=3.5$
And finaly let's check our result:
$V=xyz=7\cdot7\cdot3.5=171.5$

4 0

3 years ago

2.5: Một người nặng 72kg ngồi trên sàn treo nặng 12kg như hình vẽ. Hỏi người đó

Ivahew [28]

Answer:

english

Explanation:

5 0

2 years ago