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Sindrei [870]
3 years ago
7

Trình bày về Thí nghiệm tán xạ Rutherford, phát hiện proton.

Physics
1 answer:
Setler [38]3 years ago
6 0

Answer:

Any other language I don't know this language

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Do Small objects exert no gravitational force ? True or false
Zanzabum

Answer:

False -    

F = G M1 M2 / R^2

So F depends on M1 and M2 and as long either is not zero there will be a gravitational force between them.

6 0
2 years ago
Which gas giant has a rotation axis so tilted that the planet rotated like a bowling ball as it orbits the sun?
Anestetic [448]
The answer to your question is OPTION B
3 0
3 years ago
A bird with a mass of 0.88 kg has a potential energy of 96 J. How high off the ground is the bird?
Ray Of Light [21]

Answer:

Data:-m=0.88kg ,g=9.8m/sec² ,P.E=96J ,h=?

Explanation:

solution ,P.E=mgh here we have to find h so h=P.E/mg ,h=96/0.88×9.8 ,h=96/8.624=11.131m and if you want to verify so just put the value of h in same formula, likewise :-P.E=mgh ,P.E=0.88×9.8×11.131=96J so we got the same value of P.E as it is given the question (verified).

5 0
3 years ago
You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds
german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

f_{beat}=\frac{1}{T}

Then the frequency for the previous period given (2sec) is

f_{beat}=\frac{1}{2}

f_{beat} = 0.5Hz

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz

<em>Hence option A is incorrect.</em>

We can do this process for 254Hz as f_1 and 258 Hz for f_2 , then

f_{beat} =|254Hz-258Hz|

f_{beat} = 4Hz

<em>Hence option B is incorrect. </em>

We can also do this process for 255Hz as  f_1 and 257 Hz for f_2 , then

f_{beat} =|255Hz-257Hz|

f_{beat} = 2Hz

<em>Hence option C is incorrect. </em>

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz

<em>Hence option D is incorrect. </em>

We can also do this process for 255.75Hz as f_1 and 256.25 Hz for f_2, then

f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz

<em>Hence option E is incorrect. </em>

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0
3 years ago
A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 =
Liula [17]
<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam. 
1)What is the force the left support exerts on the beam? 
2)What is the force the right support exerts on the beam? 
3)How much extra mass could the gymnast hold before the beam begins to tip? 
Now the gymnast (not holding any additional mass) walks directly above the right support. 

4)What is the force the left support exerts on the beam? 
5)What is the force the right support exerts on the beam?</span>
6 0
3 years ago
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