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3 years ago

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A north magnetic pole is facing another north magnetic pole with a distance x. If the distance between the poles becomes 12x, wh

at happens to the magnitude of the field energy between them?
The answer is "The field energy will increase"

1 answer:

krek1111 [17]3 years ago

4 0

Answer:

1.)The field energy will increase.

The rest of the answers:

2.)The energy increases, and the lines of force are denser

3.) It points toward the field of earths magnetic poles.

4.) l, ll, and lll only

5.) ll, lV, l, lll

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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 : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

Which of the following tools can be used to determine humidity? Select all that apply

Y_Kistochka [10]

Thermometer there's others you can use but i know that's one of them

7 0

3 years ago

Read 2 more answers

Questions 15 out of 20

Olin [163]

Answer:

D

Explanation:

7 0

3 years ago

A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3

vaieri [72.5K]

22.5 J

Explanation:

Given:

x = 3 m

$k = 5\:\text{N/m}$

The spring potential energy $PE_s$ is

$PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2$

$\:\:\:\:\:\:\:=22.5\:\text{J}$

3 0

2 years ago

A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco

Sunny_sXe [5.5K]

Answer:

(a)

$x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km$

(b) $r=41.32km$

$\alpha =24.23^o$

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies $45^o$ south of east. The y-component will be negative and the x-component will be positive.

$x_1=25cos45^o=17.68km$

$y_1=-25sin45^o=-17.68km$

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

$x_2=40cos60^o=20km$

$y_2=40sin60^o=34.64km$

Therefore, total displacements:

$x=x_1+x_2=(17.68+20)km=37.68km$

$y=y_1+y_2=(-17.68+34.64)km=16.96km$

Magnitude of displacements,

$r=\sqrt{x^2+y^2}=41.32km$

Direction,

$\alpha =tan^{-1}(\frac{y}{x})=24.23^o$

6 0

3 years ago