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yulyashka [42]
2 years ago
5

What is the best explanation for why a magnet is different from a regular piece of metal?

Physics
1 answer:
arlik [135]2 years ago
5 0

Answer:

in a magnet there is a magnetic field that draws ever mental to it

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A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga
tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

               d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₂² v₂

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{2.25^2}{3.75^2} v_2

 v_1= 0.36 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)

 32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)

 v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}

 v_2=\sqrt{16.084}

       v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)²  x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0
3 years ago
A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

8 0
3 years ago
A parallel-plate capacitor in air has a plate separation of 1.76 cm and a plate area of
Monica [59]

Answer:

Explanation:

Plate separation, d = 1.76 cm = 0.0176 m

Area of plates, A = 25 cm^2 = 0.0025 m^2

V = 255 V

(a) Capacitance of capacitor

C = \frac{\epsilon _0A}{d}

C = \frac{8.854\times 10^{-12}\times 0.0025}{0.0176}

C = 1.258 x 10^-12 F

charge is same before and after immersion as the battery is disconnected

q = C V

q = 1.258 x 10^-12 x 255 = 3.2 x 10^-10 C

(b)

Capacitance before, C = 1.258 x 10^-12 C

capacitance after, C' = k x C = 80 x 1.258 x 10^-12 = 100.64 x 10^-12 C

Where, k is the dielectric constant of water = 80

Potential difference after immersion, V' = V / k = 255 / 80 = 3.1875 V

(c) initial energy,

U = \frac{q^{2}}{2C}

U = \frac{(3.2\times 10^{-10})^{2}}{2\times 1.258\times 10^{-12 }}=4.07\times 10^{-8}J

Final energy

U' = \frac{q^{2}}{2C'}

U' = \frac{(3.2\times 10^{-10})^{2}}{2\times 100.64\times 10^{-12}}=5.08\times 10^{-10}J

6 0
3 years ago
Two identical bullets are used. Both are released at the same height - one fired out of a gun, the other is dropped. Ignoring ai
irina [24]

Answer:

Both bullets will hit the ground at the same time.

Explanation:

Let's only analyze the vertical problem.

Any object that is not in the floor or resting in some site is being affected by the gravitational force (remember that we are ignoring air resistance)

Then the acceleration of this object will be equal to the gravitational acceleration:

a = -9.8m/s^2

Where the minus sign is because this acceleration goes down.

To get the velocity equation we need to integrate over time, we will get:

v(t) = ( -9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity.

To get the position equation we need to integrate over time again, we will get:

p(t) = (1/2)*( -9.8m/s^2)*t^2 + v0*t + H

Where H is the initial height.

p(t) = (-4.9 m/s^2)*t^2 + v0*t + H

The object will hit the ground when p(t) = 0

Then we need to solve for t the next equation:

(-4.9 m/s^2)*t^2 + v0*t + H = 0

Notice that the only things we need to know are:

H = initial height (we know that is the same for both bullets)

v0 = initial vertical velocity (also is the same for both bullets)

Notice that the horizontal velocity does not affect this equation, then we will get the same value of t for the dropped bullet and for the fired bullet.

This means that both bullets will hit the ground at the same time.

8 0
3 years ago
the body Will remain at rest or move at constant velocity unless acted by external net force are unbalance force​
Vikentia [17]

Answer:

law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. This is also known as the law of inertia. Inertia is the tendency of an object to remain at rest or remain in motion.

can u give me

your crown to me please brainliest me id you like my answer

4 0
2 years ago
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