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Finger [1]
3 years ago
9

what polarity must the red tip of the compass needle be since it points toward the south pole of the magnet?

Physics
1 answer:
Marrrta [24]3 years ago
8 0

Answer: positive

Explanation: because opposites attract and north=positive and south=negative the answer should be positive.

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The magnet below is cut in half. what will be the result?
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the correct would be c

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a magnet will always have a north and a south so if you cut the magnet in half the north part would have a south on the opposite end and vice versa .

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Which of the following best describes a chemical equation that demonstrates the law of conservation of matter?
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3 years ago
A concrete block of mass 35kg is pulled along a horizontal floor with the aid of a rope inclined at an angle of 30⁰ to the horiz
TEA [102]

Answer:

C. 211.5 N

Explanation:

We included a free body diagram representation of the concrete block. The equation of equilibrium in the x-direction:

\Sigma F_{x} = P\cdot \cos \alpha -\mu_{k}\cdot N = m\cdot a (1)

\Sigma F_{y} = N + P\cdot \sin \alpha -m\cdot g = 0 (2)

Where:

P - External force exerted on the block, measured in newtons.

N - Normal force from the ground to the block, measured in newtons.

\alpha - Direction of the external force, measured in sexagesimal degrees.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

m - Mass of the concrete block, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

a - Net acceleration of the concrete block, measured in meters per square second.

The minimum force to move block occurs when net force is zero, from (2) we get that the normal force is:

N = m\cdot g -P\cdot \sin \alpha

And by applying this formula in (1), we get that the external force is:

P\cdot \cos \alpha - \mu_{k}\cdot (m\cdot g -P\cdot \sin \alpha) = m\cdot a

P\cdot \cos \alpha -\mu_{k}\cdot m\cdot g +\mu_{k}\cdot P\cdot \sin \alpha = 0

P\cdot (\cos \alpha +\mu_{k}\cdot \sin \alpha)=\mu_{k}\cdot m\cdot g

P = \frac{\mu_{k}\cdot m\cdot g}{\cos \alpha +\mu_{k}\cdot \sin \alpha}

If we know that \mu_{k} = 0.75, m = 35\,kg, g = 10\,\frac{m}{s^{2}} and \alpha = 30^{\circ}. then the external force is:

P = \frac{(0.75)\cdot (35\,kg)\cdot \left(10\,\frac{m}{s} \right)}{\cos 30^{\circ}+0.75\cdot \sin 30^{\circ}}

P \approx 211.518\,N

The correct answer is C.

6 0
3 years ago
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