Two of the three tuning forks have known frequencies. When a 510 Hz. fork and the unknown fork are struck together, four beats per second are heard. When a 505 Hz. fork is struck with the unknown fork, beats are produced that are too numerous to count accurately. What is the frequency of the unknown fork?
Answers
Therefore the unknown is either 514 or 506.
If 505 and 506 were struck together, the diff would be 1 Hz
If 505 and 514 were struck toghether, the diff would be 9 Hz which would be difficult to count accurately,
Therefore the unknown is 514
Answer:
514Hz
Explanation:
510 - unknown = 4Hz
unknown = 514 or 506
505 - unknown = numerous
unknown is most likely 514 because 1 beat that wouldve resulted from the 506 Hz isn't numerous.
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Answers
\[ F_{\text{net}} = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Where:
- \(G\) is the gravitational constant (\(6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\))
- \(m_1\) and \(m_2\) are the masses of the two objects (160 kg and 460 kg)
- \(r\) is the separation between the objects (0.440 m)
Plugging in the values:
\[ F_{\text{net}} = \frac{(6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2) \cdot (160 \, \text{kg}) \cdot (460 \, \text{kg})}{(0.440 \, \text{m})^2} \]
\[ F_{\text{net}} \approx 5.1 \, \text{N} \]
The direction of the net force is along the line connecting the two masses.
(b) To find the position where a 51.0 kg object can be placed so as to experience a net force of zero, we can set up an equation using the gravitational forces from each mass. Let's assume the 51.0 kg object is placed at a distance \(x\) from the 460 kg mass. The net force can be zero if the gravitational force from the 160 kg mass is equal and opposite to the gravitational force from the 460 kg mass:
\[ \frac{G \cdot m_{\text{obj}} \cdot m_{160}}{x^2} = \frac{G \cdot m_{\text{obj}} \cdot m_{460}}{(0.440 - x)^2} \]
Where:
- \(m_{\text{obj}}\) is the mass of the object (51.0 kg)
- \(m_{160}\) is the mass of the 160 kg object
- \(m_{460}\) is the mass of the 460 kg object
- \(x\) is the distance from the 460 kg mass
Solving for \(x\):
\[ \frac{m_{160}}{x^2} = \frac{m_{460}}{(0.440 - x)^2} \]
\[ m_{160} \cdot (0.440 - x)^2 = m_{460} \cdot x^2 \]
Solving this quadratic equation will give you the position \(x\) where the net force on the 51.0 kg object is zero.
Answers
Answer:
The apparent weight is the weight of the body minus the weight of the liquid displaced. The body will float only when both the weights are same. In this case, the given body of weight W is floating and hence the apparent weight is zero.
Answer:
Zero
Explanation:
The apparent weight is the weight of the body minus the weight of the liquid displaced. The body will float only when both the weights are same. In this case, the given body of weight W is floating and hence the apparent weight is zero.
Answers
b) They are mostly negative.
c) They are close to zero.
d) They are mostly positive.
Answers
Answer: The correct option is (c).
Explanation:
The electric force exists between the charged particles only.
The electric force between the negatively or positively charged particles is attractive in nature while the electric force between the negative and the positive charged particles is repulsive in nature.
Matter consists of atoms. Atom consists of electrons, protons and neutrons. Only, electrons can move. If the atom loses electrons then matter gets negatively and positively charge respectively.
In the given problem, the matter is mostly neutral on a large scale. On a large scale, the electric forces are close to zero.
Therefore, the correct answer is (c).
Answers
Kinetic energy = (1/2) (mass) (speed)²
= (1/2) (65 kg) (5 m/s)²
= (32.5 kg) (25 m²/s²)
= 812.5 kg-m²/s² = 812.5 joules .
Answers
Answer: The slope of a line on a distance-time graph is- speed of the object.
The slope of a line on a graph refers to rate of change of variable that is presented on Y axis with respect to the variable that is presented on X axis.
For a distance time graph, distance is presented on Y axis and time on the X axis.
As we know that
Therefore, the slope of a line on a distance-time graph represents speed of the object.
Slope of any given curve is defined as
Rate of change in quantity on Y axis with respect to the quantity on x axis.
Here on Y axis if we plot distance and on X axis if time is plotted then
Slope = [tex] \frac{ds}{dt}[/tex}
above expression is rate of change in distance with time which shows apped of object