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Archy [21]
3 years ago
7

The number of waves passing a fixed point per second is known as ______________.

Physics
1 answer:
igomit [66]3 years ago
5 0
<span>The number of waves passing a fixed point per second is known as frequency. The higher the number is, the greater is the frequency of the wave. The SI unit for wave frequency is the hertz (Hz).</span>
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Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius?
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<span>Jun 16, 2012 - Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius? –73°C 27°C 327°C 673°C.</span><span>
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3 0
3 years ago
Read 2 more answers
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

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
14. Which of the following is an example of
Murljashka [212]

Answer:

none of the above or screwdriver

4 0
2 years ago
Where on the physical activity pyramid do lifestyle activities belong?
Virty [35]

Answer:

The physical activity pyramid helps in organizing the person's daily life routine. It separated the exercises into different levels and helps the person to inherit a fit and healthy life style.

5 0
2 years ago
A fan blade, initially at rest, rotates with a constant acceleration of 0.029 rad/s2. What is the time interval required for it
LenKa [72]

Answer:

The time interval is  t = 21.30 \ s

Explanation:

From the question we are told that

    The constant acceleration is \alpha  = 0.029 \ rad / s^2

    The displacement is  \theta  =  6.58 \ rad

     

According to the second equation of motion we have that

    \theta  =  w_i* t  +  \frac{1}{2} *  \alpha  t^2

given that the blade started from rest

     w_i which is the initial angular velocity is 0

 So  

       \theta  = 0 +  \frac{1}{2} *  \alpha  t^2

 =>  t = \sqrt{ \frac{2 * \theta }{\alpha } }

substituting values  

=>    t = \sqrt{ \frac{2 * 6.58 }{0.029 } }

=>    t = 21.30 \ s

6 0
3 years ago
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