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Damm [24]
3 years ago
6

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a

what is the initial acceleration of the proton after it is released
Physics
1 answer:
mihalych1998 [28]3 years ago
8 0
<span>We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
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Answer:

2.5m/s^2

Explanation:

Step one:

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v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a

divide both sides by 40

a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2

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A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?
Montano1993 [528]

Net Force = (mass) x (acceleration)  (Newton #2)

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Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

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Axis y

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Body B

Vertical axis

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In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

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We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

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     m_A = M

     m_B = M

     a = M / (1 + 1) M g

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Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

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We seek tension for this case

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Let's look for the relationship between the tensions of the two cases

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   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
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