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zepelin [54]
3 years ago
5

Protons and neutrons grouped in a specific pattern

Physics
1 answer:
alexgriva [62]3 years ago
5 0
Answer b protons and electrons
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If the velocity of a proton is straight up (thumb pointing up) then RHR2 shows that the force points to the left. What would the
diamong [38]

Answer:

a) to the right

b) up

c) down

d) there will be no force on the proton

e) there will be no force on the proton

Explanation:

The complete question is

If the velocity of a proton is straight up(thumb pointing up) then RHR2 (right hand rule) shows that the force points to the left. What would the direction of the force be if the velocity were a)down, b)to the right, c)to the left, d)into the page, and e)out of the page?

The right hand rule for a positive charge states that...

Hold the thumb of the right hand at right angle to the rest of the fingers, and the rest of the fingers parallel to one another, and pointing away from the body. If the thumb shows the velocity of a positive charge in a magnetic field, and the fingers all point in the direction of the magnetic field, then the palm will push in the direction of the force on the positive charge (proton).

From this, we can deduce from the original statement about this proton that the direction of the field is into the screen of this computer. with that field direction held constant, we can work out that

a) if the thumb point down, the force will be to the right

b) if the thumb points to the right, the force will be upwards

c)  if the thumb points to the left, the force is downwards

d)  if the thumb points into the page, then there will be no force on the proton since the proton must travel perpendicularly to the magnetic field for a force to be induced on it

e) explanation is the same as foe option d.

6 0
3 years ago
A balloon is filled with helium at sea level. The volume was measured to be 3.1 liters at a pressure of 0.97 atm. If the balloon
stira [4]

Answer:

4.9L hope this helped!

8 0
2 years ago
what law of motion that states that an object will remain at rest or move in a straight line at a constant speed unless it is ac
prohojiy [21]

That's Newton's 1st law of motion, sometimes called the inertia law.
7 0
2 years ago
In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?
kirza4 [7]

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰  (angle with respect x-axis)

Explanation:

E(t)  Electric Field is a vector, so we need to determine module and direction

E(t)  =  E(q₁)  + E (q₂)  Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂  respectively.

E(q₁) = K * q₁/ (d₁)²         K = 9 *10⁹   [N*m²/C²]    d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225      [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) =  9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9        tanα =  2,395      α = 67,5⁰

The angle of the vector electric field with the x-axis

3 0
3 years ago
Read 2 more answers
Topic Gravitational force amd firld strength.. help me please
I am Lyosha [343]

The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

x \approx 6.74\,\mathrm m

5 0
2 years ago
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