3) Magnets are used to separate steel cans from aluminum cans in recycling plants. How can

Physics

1 answer:

Elden [556K]3 years ago

4 0

Answer:

Since steel contains iron (a magnetic metal), the magnets will attract the steel cans since aluminum is not magnetic. This is used to separate the steel cans from the aluminum cans so they can be recycled separately.

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Cozebilir misiniz !!!

GREYUIT [131]

Answer:

Ya inan bana çok özür dilerim cevabı bilmiyorum ama ilk defa bugün indirdim ve hiç türk yoktu selam vermek istedim selam jwlwpskwks herkes yabancı burda neden?

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2 years ago

A 0.500 kg rock is whirled in a vertical circle of a radius 0.60 m . the velocity of the rock at the bottom of the swing is 4.0

OLEGan [10]

Explanation:

Centripetal acceleration is:

a = v² / r

a = (4.0 m/s)² / 0.60 m

a = 26.6 m/s²

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3 years ago

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld

lesantik [10]

Answer: The magnitude of the proton's acceleration is 0.748 ×10^14 m/s²

Explanation:
the velocity ,v, of ththe proton = 5.02×10^6 m/s
Magnitude , B , of the magnetic field = 0.180 T

First , we need to find the magnitude of the Force on the proton. This is given by the relation :
F = q(v x B) = qvBsinθ

where 'q' is the charge of proton , q= 1.6×10^-19 C
θ is the angle the proton makes with the direction of the magnetic field

Putting the respective values of v, B ,θ in the above equation, we get:

F = (1.6×10^-19 C)(5.02×10^6 m/s)(0.180T) sin60°
∴ F = 1.25 ×10^-13 N

Now , from Newton's second law we know that ,
F=m×a

∴ a = F/m

Mass of a proton = 1.67×10^27 kg
a= 1.25 × 10^-13 N / 1.67 × 10^27 kg

a= 0.748 × 10^14 m/s² =acceleration of the proton

(To know more about Magnetic Fields : brainly.com/question/9095546)

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1 year ago

A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w

tatyana61 [14]

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s$
And so, the kinetic energy of the object is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s$
And so the new kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

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2 years ago

What is the smallest particle into which an element can be divided and still be the same substance?

shepuryov [24]

Answer:

atom

Explanation:

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3 years ago

Read 2 more answers