The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
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Answer:
Explanation:
a ) speed of passenger = circumference / time
= 2π R / Time
= 2 x 3.14 x 50 / 60
= 5.23 m /s
b )
centrifugal force = m v² /R
= (882 /9.8 ) x 5.23² / 50
= 77.47 N
Apparent weight at the highest point
real weight - centrifugal force
= 882 - 77.47
= 804.53 N
Apparent weight at the lowest point
real weight + centrifugal force
= 882 +77.47
= 959.47 N
c ) if the passenger’s apparent weight at the highest point were zero
centrifugal force = weight
mv² /R = mg
v² = gR
= 9.8 X 50
v = 22.13 m /s
d )
apparent weight
mg - mv² / R
= 882 - (882 / 9.8 )x 22.13²/50
= 882 + 882
= 1764 N
=
Mechanical wave shows dual nature
Answer:
Decreases to half.
Explanation:
From the question given above, the following data were obtained:
Initial mass (m₁) = m
Initial force (F₁) = F
Initial acceleration (a₁) =?
Final mass (m₂) = ½m
Final force (F₂) = ¼F
Final acceleration (a₂) =?
Next, we shall determine a₁. This can be obtained as follow:
F₁ = m₁a₁
F = ma₁
Divide both side by m
a₁ = F / m
Next, we shall determine a₂.
F₂ = m₂a₂
¼F = ½ma₂
2F = 4ma₂
Divide both side by 4m
a₂ = 2F / 4m
a₂ = F / 2m
Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:
a₁ = F / m
a₂ = F / 2m
a₂ : a₁ = a₂ / a₁
a₂ / a₁ = F/2m ÷ F/m
a₂ / a₁ = F/2m × m/F
a₂ / a₁ = ½
Cross multiply
a₂ = ½a₁
From the illustrations made above, the acceleration of the car will decrease to half the original acceleration
“a point at which rays of light, heat, or other radiation meet after being refracted or reflected.” Meaning multiple light rays or heat (and other forms of radiation) are all being refracted or reflecting to a certain point
