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3 years ago

A 7- kg object is accelerating to the right at 2 km/s^2. What is the magnitude of the rightward net force acting on it?

Physics

1 answer:

Andreas93 [3]3 years ago

6 0

Answer:

Net force is 14 kN.

Explanation:

Given:

The mass of the object is, $m=7\textrm{ kg}$

The acceleration of the object is, $a=2\textrm{ }km/s^{2}$

According to Newton's second law of motion,

Net force is given as the product of mass and acceleration of the object.

Therefore, the net force acting on the object is given as,

$F=ma\\F=7\times 2=14\textrm{ kN}$

Therefore, the magnitude of the rightward net force acting on it is 14 kN.

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Answer:

Option B. 3.0×10¯¹¹ F.

Explanation:

The following data were obtained from the question:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:

Capacitance (C) = Charge (Q) / Potential difference (V)

C = Q/V

With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

C = Q/V

C = 3.0×10¯⁹ / 100

C = 3.0×10¯¹¹ F.

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3 years ago

3. The center of mass (or center of gravity) of a two-particle system is at the origin. One particle

nika2105 [10]

Answer:

B) (-2.0 m, 0.0 m)

Explanation:

Given:

Mass of particle 1 is, $m_1=2.0\ kg$

Mass of particle 2 is, $m_2=3.0\ kg$

Position of center of mass is, $(x_{cm},y_{cm})=(0,0)$

Position of particle 1 is, $(x_1,y_1)=(3.0\ m,0.0\ m)$

Position of particle 2 is, $(x_2,y_2)=(?\ m,?\ m)$

We know that, the x-coordinate of center of mass of two particles is given as:

$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 3)+(3.0\times x_2)}{2.0+3.0}\\\\0=6+3x_2\\\\3x_2=-6\\\\x_2=\frac{-6}{3}=-2.0\ m$

We know that, the y-coordinate of center of mass of two particles is given as:

$y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 0)+(3.0\times y_2)}{2.0+3.0}\\\\0=0+3y_2\\\\3y_2=0\\\\y_2=\frac{0}{3}=0.0\ m$

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