What is the instantaneous velocity v of the particle at t=10.0s?

Physics

2 answers:

algol [13]3 years ago

6 0

The instantenous velocity is just the slope of the graph at a certain instant. Since the graph is a straight line, its instantenous velocity is uniform through out. v = dx / dt = (40 - 10) / (50 - 0) = 0.6 m/s.

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BlackZzzverrR [31]3 years ago

5 0

The instantaneous velocity of the particle is 0.600 m/s. Instantaneous velocity is the velocity of an object at any given time in its path.

<h3>FURTHER EXPLANATION</h3>

The instantaneous velocity of an object can be determined graphically in a position - time graph.

If the velocity of the object is constant, the graph in a position - time graph will be a straight line and the instantaneous velocity will be equal to the slope or gradient of the line.
If the velocity of the object is changing, the graph will be a curve and the instantaneous velocity can be determined from the slope of the curved line at any given moment.

To get the slope of a line, get the ratio of the rise and the run.

In the given problem, the velocity of the object is constant because its position - time graph shows a straight line. As such, the instantaneous velocity will be equal to the average velocity.

To solve for the average velocity, use the equation below:

$average \ velocity \ = \frac{displacement \ ( \Delta y)}{ time \ (\Delta x)}$

Using data from the graph,

$average \ velocity \ = \frac{40\ m - 10 \ m}{50 \ s \ - 0 \ s} \\average \ velocity \ = \frac{30 \ m}{50 \ s}\\ \\ \boxed {average \ velocity = 0.6 \frac{m}{s}}\\$

The instantaneous velocity expressed with 3 significant figures like the given in the problem, therefore, is 0.600 m/s.

<h3>LEARN MORE</h3>

Position - Time Graph brainly.com/question/11533012
Acceleration brainly.com/question/4134594
Velocity - Time Graph brainly.com/question/4710544

Keywords: instantaneous velocity, position - time graph

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Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$