B. The electric potential at the given point from the proton is 2.7 x 10¹ volts
<h3>
Electric potential of the proton</h3>
The electric potential of the proton is calculated as follows;
V = kq/r
where;
- k is Coulomb's constant
- q is charge
- r is distance
V = (9 x 10⁹ x 1.6 x 10⁻¹⁹)/(5.3 x 10⁻¹¹)
V = 27.2 volts
V = 2.7 x 10¹ volts
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Work done in compressing spring = 40.833
It simply needs a three-inch compression by applying a compressive force to a 15-inch long spring. we know that F(x)=kx. To act in displacement X and the poles' values cannot be substituted. Consequently, there will be 4. The spring constant K is therefore equal to 5, so. Three fit within this slot. We consequently derive the compressing spring constant K from this. , which is 5, 3. Now change this formula to include the value of the compressing spring constant. Four f will therefore equal 5.3 x and vice versa. Integration of four factors equals W.
In order to ensure the limit and further all of this, we currently have the range 0–7. Thus, this compressing spring task is completed. W is equal to 40.833
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Tons of stuff!
Footpath erosion, increased usage of travel vehicles, construction of hotels and other attractions, ect ect.
Answer:
3.1 s
Explanation:
from the question we are given the the following:
Weight of books (Fb) = 305 N
Push force (Fp) = 516 N
distance (s) = 5.82 m
Angle of force exerted = 32 degrees
acceleration due to gravity (g) = 9.81 m/s^{2}
coefficient of friction Uk= 0.59
time (t) = ?
mass of the book (m) = weight / (g) = 305 / 9.8 = 31 kg
lets first get the net force from the summation of vertical forces
Fnet = Fpush + Fgravity
F net = 516 + 305sin32
F net = 677.6 N
now lets get the acceleration from the summation of the horizontal forces
Fpcos32 - friction force = m x a
Fpcos32 - (Uk x F net) = m x a
516cos32 - (0.59 x 677.6) = 31 x a
37.8 = 31a
a = 1.22m/s^{2}
now that we have our acceleration we can get the time from the equation of motion
s = ut + o.5at^{2}
u ( initial velocity ) = 0 because the box was initially at rest
5.82 = (0 x t ) + (0.5 x 1.22 x t^{2})
5.82 = 0.61t^{2}
t = 3.1 s
Answer:
The magnetic flux through surface is 
Wb
Explanation:
Given :
Magnitude of magnetic field 
T
Radius of circle 
m
Angle between field and surface normal 
25°
From the formula of flux,
Where angle between magnetic field line and surface normal, 
area of circular surface.
Magnetic flux is given by,
Wb
Therefore, the magnetic flux through surface is Wb
