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3 years ago

14

A student designs an electromagnet, as shown in the picture. The electromagnet is only able to pick up 1 paper clip. List 2 modi

fications he could make to his electromagnet in order to enable it to pick up more paper clips.

1 answer:

Lerok [7]3 years ago

6 0

-- Change the battery to one with higher voltage.

-- Wrap more turns of wire around the spike.

-- If the spike is made of anything else but iron, replace it with a pure iron one.

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Brainliest for correct answer :))

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Choose the 2nd Option.....

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Pls help i’ll give brainliest if you give a correct answer!!

arsen [322]

Answer:

C. The distance traveled by an object at a certain velocity.

Explanation:

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Do Number 8<br> Please respond fast<br> WILL MARK BRAINLIEST<br> (THIS INVOLVES INCLINED PLANES)

jasenka [17]

Answer:

We need to apply a force of magnitude 32.13 N

Explanation:

The forces acting on this crate are:

1) the crate's weight (which we can decompose in two components: one parallel to the incline, and the other one perpendicular to it) The parallel component of the weight is the product of the mass, times g (9.8 m/s^2) times sin(35). This component is pointing down the incline, and let's call it the "x" component. Its magnitude is given by:

$w_x=m\,\,g\,\,sin(35^o)$

The magnitude of the perpendicular component of the weight (let's call it the y-component) is given by the expression:

$w_y=m\,\,g\,\,cos(35^o)$

2) The force of friction between the crate and the surface of the incline (acting parallel to the incline and opposite the direction of the parallel component of the weight ( $w_x$ )

3) The extra force we need to apply perpendicular to the incline and towards it (lets call it "E") so the crate doesn't slide down.

4) The normal force that the incline applies on the crate as reaction. This force is pointing away from the incline.

For the block not to move (slide down the incline), we need that the parallel component of the weight equals the force of friction with the surface of the incline. Let's call this force of friction $f_s$ , and recall that it is defined as the product of the normal force times the coefficient of friction (given as 0.3 in value). This in equation form becomes:

$f_s=m\,g\,sin(35^o)\\\mu\,\,n=m\,g\,sin(35^o)\\n=\frac{m\,g\,sin(35^o)}{\mu} \\n=\frac{3\,*\,9.8\,\,sin(35^o)}{\0.3}\\n=56.21 \,\,N$

We solved for the needed magnitude of the normal force in order to keep the crate from sliding.

Now we study the forces vertical to the incline, which should also be balanced since the crate is not moving in this direction. We add all the forces acting towards the incline (the perpendicular component of the crate's weight, and the extra force "E"), and make them equal to the only force coming outwards (the normal force):

$n=E+m\,\,g\,\,cos(35^o)$

and we can solve for the magnitude of the extra force we need to apply by replacing all other known values:

$n=E+m\,\,g\,\,cos(35^o)\\56.21\,N=E+3\,*\,9.8\,\,cos(35^o)\,N\\E= [56.21-3\,*\,9.8\,\,cos(35^o)]\,N\\E=32.13\,N$

So this is the magnitude of the force we need to apply in order to keep the crate from sliding down the incline.

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3 years ago

At some instant, a particle traveling in a horizontal circular path of radius 7.90 m has a total acceleration with a magnitude o

DochEvi [55]

Answer:

a) Speed of the particle at this instant

v = 8.43 m/s

b) Speed of the particle at (1/8) revolution later

v = 14.83 m/s

Explanation:

We apply the equations of circular motion uniformly accelerated :

$(a_{T}) ^{2} = (a_{n} )^{2} +(a_{t} )^{2}$ Formula (1)

$a_{n} = \frac{v^{2} }{r}$ Formula (2)

$a_{t} = \alpha *r$ Formula (3)

v= ω*r Formula (4)

ω² = ω₀² + 2*α*θ Formula (5)

Where:

$a_{T}$ : total acceleration, (m/s²)

$a_{n}$ : normal acceleration, (m/s²)

$a_{t}$ : tangential acceleration, (m/s²)

$\alpha$ : angular acceleration (rad/s²)

r : radius of the circular path (m)

v : tangential velocity (m/s)

ω : angular speed ( rad/s)

ω₀: initial angular speed ( rad/s)

θ : angle that the particle travels (rad)

Data:

$a_{T}$ = 15 m/s²

$a_{t}$ = 12 m/s²

r=7.90 m :radius of the circular path

Problem development

In the formula (1) :

$a_{n} = \sqrt{(a_{T})^{2} -(a_{t})^{2} }$

We replace the data

$a_{n} = \sqrt{(15)^{2} -(12)^{2}}$

$a_{n} = 9 \frac{m}{s^{2} }$

We use formula (2) to calculate v:

$9 = \frac{v^{2} }{7.9}$ Equation (1)

a)Speed of the particle at this instant

in the equation (1):

$v=\sqrt{9*7.9} = 8.43 \frac{m}{s}$

b)Speed of the particle at (1/8) revolution later

We know the following data:

θ =(1/8) revolution=( 1/8) *2π= π/4

$a_{t}$ = 12 m/s²

v₀= 8.43 m/s

r=7.9 m

We use formula (3) to calculate α

$12 = \alpha *7.90$

$\alpha =\frac{12}{7.9} = 1.52 \frac{rad}{s^{2} }$

We use formula (4) to calculate ω₀

v₀= ω₀ *r

8.43 = ω₀*7.9

ω₀ = 8.43/7.9 = 1.067 rad/s

We use formula (5) to calculate ω

ω² = ω₀² + 2*α*θ

ω²= (1.067)² + 2*1.52*π/4

ω² =3.526

ω = 1.87 rad/s

We use formula (4) to calculate v

v= 1.87 rad/s * 7.9m

v = 14.83 m/s : speed of the particle at (1/8) revolution later

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3 years ago

Plants are important to the balance of nature because they provide food and

Sphinxa [80]

Plants provide food and Carbon dioxide

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3 years ago