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Lesechka [4]
3 years ago
8

What is the acceleration of a 10 kg mass pushed by a 5 N force?

Physics
2 answers:
yanalaym [24]3 years ago
8 0
g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}=  \\ g=0,5 \frac{N}{kg}
professor190 [17]3 years ago
6 0
Force equals mass times acceleration. Or:
F=ma
Plug it in:
5=10a
5/10=(10a)/10
.5m/s²=a
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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha
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Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

Given:

mass of first particle, m_1=6.7\ kg

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mass of third particle, m_3=3.7\ kg

coordinate position of first particle in meters, (x_1,y_1)\equiv(4.2,0)

coordinate position of second particle in meters, (x_2,y_2)\equiv(0,2.8)

coordinate position of third particle in meters, (x_3,y_3)\equiv(0,0)

<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

F_{32}=16.05\times 10^{-11}\ N

towards positive X axis.

<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

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