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seraphim [82]
2 years ago
15

Convert time from 12-hour to 24-hour clock. ​

Physics
1 answer:
irina [24]2 years ago
3 0
18:06 is 6:06pm converted to 24 hour time.
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The particle in the atom with a negative charge is the ______<br> Answer here
NISA [10]

Answer:

Explanation:

The electron has a negative charge. Proton is positive and neutron is neutral.

4 0
3 years ago
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Calculate the length of a simple pendulum that oscillates with a frequency of 0.4Hz g=10m/s2 , ^=3.142
earnstyle [38]

Answer:

Explanation:

For simple pendulum the formula is

T=2\pi\sqrt{\frac{l}{g} }

Where T is time period , l is length and g is acceleration due to gravity .

\frac{1}{n} =2\pi\sqrt{\frac{l}{g} }

n is frequency

Putting the values

\frac{1}{.4} =2\pi\sqrt{\frac{l}{10} }

\frac{l}{10} = .1584

l = 1.584 m

4 0
3 years ago
If two bicycles of same masses move at different velocities,it will be easier to stop the bicycle that is moving at lower veloci
s344n2d4d5 [400]

Answer:

It is easier to stop the bicycle moving at a lower velocity because it will require a <em>smaller force</em> to stop it when compared to a bicycle with a higher velocity that needs a<em> bigger force.</em>

Explanation:

The question above is related to "Newton's Law of Motion." According to the <em>Third Law of Motion</em>, whenever an object exerts a force on another object <em>(action force)</em>, an equal force is exerted against it. This force is of the same magnitude but opposite direction.

When it comes to moving bicycles, the force that stops their movement is called "friction." Applying the law of motion, the higher the speed, the higher the force<em> </em>that is needed to stop it while the lower the speed, the lower the force<em> </em>that is needed to stop it.

8 0
2 years ago
What is the acceleration for an object moving with constant speed? Explain your answer
Oduvanchick [21]

Answer:

What is the acceleration of an object moving at a constant speed?

The Meaning of Constant Acceleration

The data table above show an object changing its velocity by 10 m/s in each consecutive second. This is referred to as a constant acceleration since the velocity is changing by a constant amount each second.

3 0
3 years ago
Read 2 more answers
Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o
Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.  

\lambda_{proton} = 6.05x10^{-14}m < \lambda_{electron} = 1.10x10^{-10}m

Part B:

The proton has a smaller wavelength than the electron.

\lambda_{proton} = 1.29x10^{-13}m < \lambda_{electron} = 5.525x10^{-12}m

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

\lambda = \frac{h}{p} (1)

Where h is the Planck's constant and p is the momentum.

\lambda = \frac{h}{mv} (2)

Part A

Case for the electron:

\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

But J = Kg.m^{2}/s^{2}

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

\lambda = 1.10x10^{-10}m

Case for the proton:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}

\lambda = 6.05x10^{-14}m

Hence, the proton has a smaller wavelength than the electron.  

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

KE = \frac{1}{2}mv^{2}

2KE = mv^{2}

v^{2} = \frac{2KE}{m}

v = \sqrt{\frac{2KE}{m}}  (3)

Case for the electron:

v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}

but 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}

v = 1.316x10^{8}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}    

\lambda = 5.525x10^{-12}m

Case for the proton :

v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}

But 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}

v = 3.07x10^{6}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}

\lambda = 1.29x10^{-13}m    

Hence, the proton has a smaller wavelength than the electron.

7 0
3 years ago
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