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2 years ago

14

If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric fiel

d point

1 answer:

Grace [21]2 years ago

3 0

Answer:

There will be no change in the direction of the electric field .

Explanation:

The direction will remain the same because the sign of the charges has no effect on it.

When one replaces the conducting cube with one that has positive charge carriers there will be no change in the direction of the field as there is no defined relationship between the direction of the electric field and sign of the charge.

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A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

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2 years ago

Which of the following is NOT an advantage of using a wheel and axle?

kakasveta [241]

Answer:

the first one

Explanation:

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2 years ago

Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm

Gre4nikov [31]

Here since both children and merry go round is our system and there is no torque acting on this system

So we will use angular momentum conservation in this

$I_1\omega_1 = I_2\omega_2$

now here we have

$I_1 = \frac{MR^2}{2} + m_1R^2 + m_2R^2$

$I_1 = \frac{100(1.60)^2}{2} + (22 + 28)(1.60)^2$

$I_1 = 256$

Now when children come to the position of half radius

then we will have

$I_2 = \frac{MR^2}{2} + m_1(\frac{R}{2})^2 + m_2(\frac{R}{2})^2$

$I_2 = \frac{100(1.6)^2}{2} + (28 + 22)(0.8)^2$

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now from above equation we have

$256 (20.0 rpm) = 160(\omega_2)$

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<span>Chemical and kinetic energy...</span>

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Answer:

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