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GarryVolchara [31]
3 years ago
10

If the period of a clock signal is 500 ps, the frequency is:_____

Physics
1 answer:
NNADVOKAT [17]3 years ago
4 0

Answer:

The frequency of the signal is 2 GHz

Explanation:

Given;

period of the clock signal, T = 500 ps = 500 x 10⁻¹² s

the frequency of the signal is given by;

F= \frac{1}{T}\\\\F = \frac{1}{500*10^{-12}}\\\\F = 2*10^{9} \ Hz

F = 2 GHz

Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz

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Brilliant_brown [7]

Answer:

Explanation:

x

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3 years ago
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A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.
Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

a=\frac{v^{2} }{r}, where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

a=\frac{3.5^{2} }{8.5} \\a=1.4

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

7 0
3 years ago
What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?
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Answer:

there are 25 kg objective travelling at 2m/s to the right.

4 0
2 years ago
A car drives 100 km north, 50 km east, then 10 km south. What is the car's displavement?​
Over [174]

Answer:

\sqrt{ ({40}^{2} }  +  {50}^{2} ) =  \sqrt{(1600 + 2500)}  =  \sqrt{4100}  = 64.03

6 0
3 years ago
g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the
sdas [7]

Answer:

The vertical distance is  d = \frac{2}{k} *[mg + f]

Explanation:

From the question we are told that

   The mass of the cylinder is  m

    The kinetic frictional force is  f

Generally from the work energy theorem

    E  =  P +  W_f

Here E the the energy of the spring which is increasing and this is mathematically represented as

       E =  \frac{1}{2} * k  *  d^2

Here k is the spring constant

        P is the potential energy of the cylinder which is mathematically represented as

     P  = mgd

And

     W_f  is the workdone by friction which is mathematically represented as

      W_f  =  f *  d

So

    \frac{1}{2} * k  *  d^2 =  mgd +  f *  d

=>    \frac{1}{2} * k  *  d^2 =  d[mg +  f    ]

=>  \frac{1}{2} * k  *  d =  [mg +  f    ]

=> d = \frac{2}{k} *[mg + f]

5 0
3 years ago
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