Example Question - gravitational force
Here are examples of questions we've helped users solve.
Calculating Potential Energy on an Inclined Plane
<p>La energía potencial se calcula usando la fórmula:</p> <p>E_p = m \cdot g \cdot h</p> <p>Donde:</p> <p>m = 10 \, \text{kg}, \quad g = 10 \, \text{m/s}^2, \quad h = 5 \, \text{m}</p> <p>Entonces, sustituyendo los valores:</p> <p>E_p = 10 \cdot 10 \cdot 5 = 500 \, \text{J}</p>
Calculating Gravitational Force Between Celestial Bodies
<p>To solve for the gravitational force between the Sun and Proxima Centauri, we can use Newton's law of gravitation given by \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \) where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two bodies, and \( r \) is the distance between the centers of the two masses.</p> <p>\( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) (Gravitational constant)</p> <p>\( m_1 = 2.00 \times 10^{30} \, \text{kg} \) (Mass of the Sun)</p> <p>\( m_2 = 2.40 \times 10^{29} \, \text{kg} \) (Mass of Proxima Centauri)</p> <p>\( r = 4.00 \times 10^{13} \, \text{km} \) (Distance between Sun and Proxima Centauri)</p> <p>We need to convert \( r \) from kilometers to meters:</p> <p>\( r = 4.00 \times 10^{13} \times 10^3 \, \text{m} \) (since \( 1 \, \text{km} = 10^3 \, \text{m} \))</p> <p>Now we substitute the values into the equation:</p> <p>\( F = \frac{(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \cdot (2.00 \times 10^{30} \, \text{kg}) \cdot (2.40 \times 10^{29} \, \text{kg})}{(4.00 \times 10^{16} \, \text{m})^2} \)</p> <p>\( F = \frac{(6.674 \times 10^{-11}) \cdot (2.00 \times 10^{30}) \cdot (2.40 \times 10^{29})}{16.00 \times 10^{32}} \)</p> <p>\( F = \frac{(6.674 \times 2.40) \cdot 10^{48}}{16.00 \times 10^{32}} \)</p> <p>\( F = \frac{16.0184 \times 10^{48}}{16.00 \times 10^{32}} \)</p> <p>\( F = 1.00115 \times 10^{16} \, \text{N} \)</p> <p>Note: The final answer has been rounded to five significant figures.</p>
Calculation of Pressure Exerted by a Tourist on Snow
<p>The pressure \( P \) exerted by an object is given by the formula \( P = \frac{F}{A} \), where \( F \) is the force and \( A \) is the area over which the force is distributed.</p> <p>Here, the gravitational force \( F \) exerted by the tourist is equivalent to their weight, which can be calculated by the mass \( m \) times the gravitational acceleration \( g \), \( F = m \cdot g \).</p> <p>The tourist's mass \( m \) is \( 60 \) kg, and the gravitational acceleration \( g \) is given as \( 10 \) N/kg. Thus:</p> <p>\[ F = m \cdot g = 60 \, \text{kg} \cdot 10 \, \text{N/kg} = 600 \, \text{N} \]</p> <p>The total area \( A \) of the shoes in contact with the snow is \( 0.6 \) m\(^2\).</p> <p>So the pressure \( P \) is:</p> <p>\[ P = \frac{F}{A} = \frac{600 \, \text{N}}{0.6 \, \text{m}^2} = 1000 \, \text{Pa} \]</p> <p>Therefore, the answer is \( D. 1000 \) Pa.</p>
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com