Answer:
Correct Answer: B. Beaker A will be 72 °C and beaker B will be 28 °C.
This one is actually right!
Which statement supports Newton’s first law of motion?
Answer:
Newton's first law of motion is that an object at rest will stay at rest, unless acted upon by an external force.
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The "external force" can be anything, from a gust of wind, to a person moving the object. His first law also corresponds with the definition of inertia.
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A Van de Graaff generator is charged so that a proton at its surface accelerates radially outward at 1.35 ✕ 1012 m/s2. Find the following. (a) the magnitude of the electric force on the proton at that instant magnitude N direction ---Select--- (b) the magnitude and direction of the electric field at the surface of the generator magnitude N/C direction ---Select---
Answer:
(a).
We know that force is
F = m a
So
F = (1.67 x 10^(-27) x (1.38 x10^12)
F = 2.3 x 10^-15 N facing the radially outward direction
(b).
Similarly Force for charge is
F = q E = m a
So relating
E = F/q = 2.3x 10^-15 /(1.6 10^ -19
E = 144.75 N/C facing the radially outward direction
The answer to the question is
(a) The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.
(b) The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.
The answer can be explained as shown below.
Given that a proton accelerates radially outward at [tex]1.35\times 10^{12}\,m/s^2[/tex].
ie; [tex]a=1.35\times 10^{12}\,m/s^2[/tex]We know the mass of a proton, [tex]m_p = 1.67\times10^{-27}\,kg[/tex].From Newton's second law we have,
[tex]F=mg[/tex]But here, [tex]m=m_p[/tex]
So, the electric force on the proton is;
[tex]F = m_p\, a= (1.67\times10^{-27}\,kg)\, \times(1.35\times 10^{12}\, m/s^2)=2.254\times10^{-15}\,N[/tex]The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.Also, we know that, in electrostatics,
[tex]F=Eq\\ \\\implies E=\frac{F}{q}[/tex]The charge of a proton is, [tex]q=1.6\times 10^{-19}\,C[/tex]Therefore, the electric field is given by,
[tex]E=\frac{2.254\times 10^{-15}N}{1.6\times 10^{-19}\,C} = 14031.25\, N/C[/tex]The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.Learn more about the electric force here:
https://brainly.com/question/8960054