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vfiekz [6]
3 years ago
5

The attractive force that acts between all objects is the (5 points)

Physics
1 answer:
Talja [164]3 years ago
7 0
The gravitational force.
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A pair of opposite electric charges of equal magnitude is called a(n)
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Answer:

Dipole

Explanation:

A pair of opposite electric charges of equal magnitude is called a(n)

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Why is an alarm clock going "off" when it actually turns on?
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The alarm is going off because the alarm has already been set to activate.
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Which of the following is moving faster?
solong [7]
The car, because logically, a car is faster and if you do the math, 300 meters per minutes is about 17 kilometers per hour, and that is lower than the car’s 36 kilometers per hour. I would show you my work but im tired man, please mark me brainliest
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Two masses, m1 & m2 are separated by a distance of 12 m between their centers. The gravitational force attraction between th
Artemon [7]

The universal law of gravitation says that the gravitational force between the two masses is

<em>F</em>₁ = <em>G</em> <em>m</em>₁ <em>m</em>₂ / <em>r</em> ² = 6.0 × 10⁻⁶ N

where <em>r</em> = 12 m. When the distance between the two masses is reduced to <em>R</em> = 4 m, i.e. 1/3 of the original distance <em>r</em>, we have

<em>F</em>₂ = <em>G</em> <em>m</em>₁ <em>m</em>₂ / <em>R </em>² = <em>G</em> <em>m</em>₁ <em>m</em>₂ / (<em>r</em>/3)² = 1/9 <em>G</em> <em>m</em>₁ <em>m</em>₂ / <em>r </em>² = 1/9 <em>F</em>₁

so that the resulting gravitational force has 1/9 the first magnitude, or

<em>F</em>₂ = 1/9 (6.0 × 10⁻⁶ N) ≈ 6.7 × 10⁻⁷ N

6 0
3 years ago
A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in
AlexFokin [52]

Answer:

Change in momentum is 2.667\times 10^{4} kg.m/s

Solution:

The momentum of any body is the product of its mass and the velocity associated with the body and is generally given by:

\vec{p} = m\vec{v}

Now, as per the question:

Mass of the car, M = 1500 kg

The velocity in the east direction, v\hat{i} = 40\hat{i} km/h

The velocity in the north direction, v\hat{j} = 50\hat{j} km/h

Now, the momentum of the car in the east direction:

p\hat{i} = mv\hat{i} = 1500\times 40\hat{i} = 60000\hat{i} kg.km/h

Now, the momentum of the car in the north direction:

p\hat{j} = mv\hat{j} = 1500\times 50\hat{j} = 75000\hat{j} kg.km/h

Change in momentum is given by:

\Delta p = p\hat{i} - p\hat{j} = 60000\hat{i} - 75000\hat{j}

Now,

|\Delta p| = |60000\hat{i} - 75000\hat{j}|

|\Delta p| = \sqrt{60000^{2} + 75000^{2}}

|\Delta p| = 96046 kg.km/hr = \frac{96046}{3.6} = 2.667\times 10^{4} kg.m/s

(Since, 1kg.km/h = \frac{1}{3.6} kg.m/s)

4 0
3 years ago
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