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2 years ago

Jack tries to place magnets on his refrigerator at home, but they won’t stick. What could be the reason?

1 answer:

5 0

The most probable reason why the magnets won't stick on the refrigerator is that the body of the refrigerator and the magnets have like poles. If both have negative or both have positive poles facing each other, they will repel. In principle, magnets are attracted to opposite poles and like poles repel.

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A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an

Sonbull [250]

Answer:

(a) k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

(b) τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = $\frac{M}{3} (a^{2} +b^{2} )$ ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k = $\frac{1}{2} Iw^{2}$ --------------------------------2

Putting equation 1 into equation 2, we have;

k = $\frac{M}{6} (a^{2} +b^{2} )w^{2}$

k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = $\frac{Mw}{3} (a^{2} +b^{2} )$

But

τ = dτ/dt = $\frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}$ ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

8 0

1 year ago

THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A

Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

5 0

2 years ago

An antibaryon composed of two antiup quarks

Sveta_85 [38]

Answer:

(2) −1 e

Explanation:

A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.

Up, charm and top quarks have $+\frac{2}{3} e$ charge where as down, strange and bottom quarks have $-\frac{1}{3}e$ charge.

The antiparticle of up quark is antiup quark and has charge $-\frac{2}{3}e$ charge.

The antiparticle of down quark is antidown quark and has charge $+\frac{1}{3}e$ charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

$2\times (-\frac{2}{3} e)+1\times (+\frac{1}{3})e=-1e$

Thus, antibaryon has -1e charge.

5 0

2 years ago

Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An

Svetradugi [14.3K]

efficiency= [useful energy transferred ÷ total energy supply]×100%

So, [5500÷10000]×100%=0.55×100

=55%

5 0

1 year ago

Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc

xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0

1 year ago