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Luda [366]
2 years ago
9

Use what you have learned in the reading to answer the following question. The point of a compass needle points to Earth’s north

pole. The point in the compass needle has–
~the same polarity as Earth's north pole.
~the opposite polarity from Earth’s north pole.
~no polarity.
~unlimited polarity.
Physics
1 answer:
DIA [1.3K]2 years ago
6 0
I would say unlimited polarity because the compass’s needle is always attracted to Earth’s north pole.

Good luck to you!
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I am going to say

C.  Energy contained in the nucleus of an atom

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3 years ago
An object travels 7.5 m/s toward the west . Under the influence of a constant net force of 5.2 kN, it comes to rest in 3.2 s. Wh
Softa [21]

Answer:

m = 2218.67 kg

Explanation:

It is given that,

Initial velocity, u = 7.5 m/s

Final speed of an object, v = 0 (at rest)

Force, F = 5.2 kN

Time, t = 3.2 s

We need to find the mass of the object. Force acting on an object is given by :

F = ma

m is mass, a is acceleration

F=\dfrac{m(v-u)}{t}\\\\m=\dfrac{Ft}{v-u}\\\\m=\dfrac{5.2\times 10^3\times 3.2}{0-7.5}\\\\m=2218.67\ kg

So, the mass of the object is 2218.67 kg

5 0
3 years ago
Why is salami better than boloni
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Ablok slides on ahorizonted sur fors which has been lubricated with heavy oil such that the blok soffers a viscous resistance th
svetoff [14.1K]

Answer:

v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}

x = \frac{2m}{c}*v_0^{\frac{3}{2}}

Explanation:

Given

f(v) =- cv^\frac{1}{2}

To start with, we begin with

F = ma

Substitute the expression for F

-cv^\frac{1}{2} = ma

-ma = cv^\frac{1}{2}

Acceleration (a) is:

a = \frac{dv}{dt}

So, the expression becomes:

m\frac{dv}{dt} = -cv^\frac{1}{2}

-----------------------------------------------------------------------------------------

Velocity (v) is:

v = \frac{dx}{dt} --- distance/time

v * \frac{dv}{dx}= \frac{dx}{dt}* \frac{dv}{dx}

v * \frac{dv}{dx}= \frac{dv}{dt}

\frac{dv}{dt} = v * \frac{dv}{dx}

--------------------------------------------------------------------------------------------

So, we have:

m\frac{dv}{dt} = -cv^\frac{1}{2}

mv * \frac{dv}{dx} = -cv^\frac{1}{2}

Divide both sides by v^\frac{1}{2}

mv^{1-\frac{1}{2}} * \frac{dv}{dx} = -c

mv^{\frac{1}{2}} * \frac{dv}{dx} = -c

Divide both sides by m

v^{\frac{1}{2}} * \frac{dv}{dx} = -\frac{c}{m}

v^{\frac{1}{2}} * dv = -\frac{c}{m} * dx

Integrate:

\int\limits^v_{v_0} {v^{\frac{1}{2}}} \, dv  = -\frac{c}{m}\int\limits^x_0 {}} \, dx

\frac{2}{3}v^{\frac{3}{2}}|\limits^v_{v_0}  = -\frac{c}{m}x|\limits^x_0

\frac{2}{3}(v^{\frac{3}{2}} - v_0^{\frac{3}{2}} ) = -\frac{cx}{m}

v^{\frac{3}{2}} - v_0^{\frac{3}{2}} = -\frac{3cx}{2m}

v^{\frac{3}{2}} = v_0^{\frac{3}{2}}-\frac{3cx}{2m}

v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}

Next, is to get the maximum velocity by distance x.

To do this, we find the derivation by x

\frac{dv}{dx} = 0

\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2}{3} - 1} * -\frac{3c}{2m} = 0

\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2-3}{3}} * -\frac{3c}{2m} = 0

\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{3c}{2m} = 0

(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{c}{m} = 0

Divide both sides by -\frac{c}{m}

(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} = 0

Take cube roots of both sides

(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{-1} = 0

v_0^{\frac{3}{2}}-\frac{3cx}{2m} = 0

\frac{3cx}{2m} = v_0^{\frac{3}{2}}

x = \frac{2m}{c}*v_0^{\frac{3}{2}}

7 0
3 years ago
Một vật không mang điện sẽ bị nhiễm điện dương khí
DochEvi [55]

Answer:

không có điện

Explanation:

7 0
3 years ago
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