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3 years ago

Five properties of magnet

Physics

1 answer:

Olenka [21]3 years ago

3 0

Answer:

1. The magnet is magnetic and can attract iron articles.

2. The magnet has magnetic poles. Each magnet has two kinds of poles: N pole and S pole. They are in pairs.

3. Temporary magnet and permanent magnet: when the ferromagnetic material is magnetized, it is easy to lose the magnetic property, which is called temporary magnet (for example: iron); when the ferromagnetic material is magnetized, it is not easy to lose the magnetic property, which is called permanent magnet (for example: steel).

4. When two magnets are close to each other, the same poles will repel and push away from each other, and the different poles will attract and stick to each other. Therefore: the same pole repels each other, the different pole attracts each other.

5. The attraction of a magnetic object is called magnetism. An object is surrounded by a magnetic material. The area affected by the magnetic force is called the magnetic field.

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A 10 kg box slides down a lane inclined at an angle θ = 30 . The plane has a friction of coefficient 0.1. The box starts from th

g100num [7]

Answer:

The distance the box traveled down the plane is 19.28 m

Explanation:

The angle of repose, α, is given by the relation;

tan⁻¹(μ) = α

tan⁻¹(0.1) = 5.7°

Therefore, we have;

M·g·sin(θ) - μ·N = M·a

Where:

M = Mass of the box = 10 kg

g = Acceleration due to gravity = 9.81 m/s²

θ = Angle of inclination of the plane = 30°°

μ = Coefficient of friction = 0.1

a = Acceleration of the box along the incline plane

N = Normal force due to the weight of the box = M·g·cos(θ)

10 × 9.81 × sin30 - 0.1 × 9.81 × cos(30) = 10 × a

48.2 = 10 ×a

a = 48.2/10 = 4.82 m/s²

The distance, s, traveled by the box is given by the relation;

s = u·t + 1/2×a·t²

Where:

u = Initial velocity = 0 m/s

t = Time of motion = 2.0 s

∴ s = 0×2 + 1/2 × 4.8 × 2² = 19.28 m

The box traveled 19.28 m down the plane.

6 0

3 years ago

What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Paul [167]

Answer:

$\Delta P=1581357.92\ Pa$

Explanation:

Given:

diameter of hose pipe, $D=0.09\ m$

diameter of nozzle, $d=0.03\ m$

volume flow rate, $\dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}$

Now, flow velocity in hose:

$v_h=\frac{\dot V}{\pi.D^2\div 4}$

$v_h=\frac{0.04\times 4}{\pi\times 0.09^2}$

$v_h=6.2876\ m.s^{-1}$

Now, flow velocity in nozzle:

$v_n=\frac{\dot V}{\pi.d^2\div 4}$

$v_n=\frac{0.04\times 4}{\pi\times 0.03^2}$

$v_n=56.5884\ m.s^{-1}$

We know the Bernoulli's equation:

$\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2$

when the two points are at same height then the eq. becomes

$\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}$

$\Delta P=\frac{\rho(v_n^2-v_h^2)}{2}$

$\Delta P=\frac{1000(56.5884^2-6.2876^2)}{2}$

$\Delta P=1581357.92\ Pa$

8 0

3 years ago

A research submarine has a 30-cm-diameter window that is 8.6cm thick. The manufacturer says the window can withstand forces up t

Triss [41]

What is missing is the density of salt water. Let's denote it ρ.
The pressure at depth d then is: 
p = p0 + ρ g d 
, where p0 is the atmospheric pressure at the surface of the sea. We assume it is equal to the 1 atm inside the sub. 

Hence the pressure difference between inside and outside of the window is: 

Δp = ρ g d 

The force on the window is this pressure difference, multiplied by the area of the circular window of diameter D. The latter equals 
A = ¼ π D². 

So 
F = p A = ¼ π D ²ρ g d. 
This should be less than Fmax. 

Hence d < Fmax / (¼ π D ²ρ g ).

6 0

3 years ago

Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the fi

laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m$

Distance from second drop is 0.83999 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m$

Distance from third drop is 0.20999 m

6 0

4 years ago

Two tiny, spherical water drops, with identical charges of −8.00 ✕ 10^(−17) C, have a center-to-center separation of 2.00 cm.

Damm [24]

Answer:

F=1.4384×10⁻¹⁹N

Explanation:

Given Data

Charge q= -8.00×10⁻¹⁷C

Distance r=2.00 cm=0.02 m

To find

Electrostatic force

Solution

The electrostatic force between between them can be calculated from Coulombs law as

$F=\frac{kq^{2} }{r^{2} }$

Substitute the given values we get

$F=\frac{(8.99*10^{9} )*(-8.00*10^{-17} )^{2} }{(0.02)^{2} }\\ F=1.4384*10^{-19} N$

7 0

3 years ago

Five properties of magnet​

Five properties of magnet