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Nutka1998 [239]
3 years ago
14

A 42 kg object is 2.5 m from a 55kg object. How much is the force of attraction between them?

Physics
1 answer:
melisa1 [442]3 years ago
8 0

Answer:

Force of attraction is 2,46*10^-8NF= G*(m1*m2/r^2)\\where \\\\G=6,67*10^-11 (N*m^2/kg^2)\\m1=42kg\\m2=55kg\\r=2,5m\\therefore\\F=6,67*10^-11*(42*55/2.5^2)\\ F=2,46*10^-8 N

Explanation:

Using the Law of Universal Gravitation, proposed by Newton.

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-- From the top of the seat to the bottom, heat is transferred by conduction.  

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Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (
lesya [120]

Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

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3 years ago
A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.
cupoosta [38]

Answer:

a) The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

\vec p = m\cdot \vec v (1)

Where:

\vec p - Vector momentum, measured in kilogram-meters per second.

m - Mass of the particle, measured in kilograms.

\vec v - Vector velocity, measured in meters per second.

If we know that m = 2.93\,kg and \vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right], then the momentum is:

\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]

\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]

The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}} (2)

\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right) (3)

Where:

\|\vec p\| - Magnitude of momentum, measured in kilogram-meters per second.

\theta - Direction of momentum, measured in sexagesimal degrees.

If we know that p_{x} = 8.731\,\frac{kg\cdot m}{s} and p_{y} = -11.661\,\frac{kg\cdot m}{s}, then the magnitude and direction of momentum are, respectively:

\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}

\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}

\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)

\theta \approx 306.823^{\circ}

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

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3 years ago
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