A)they can interbreed and produce fertile offsprings
Answer:
a) 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b) 3.466 × 10¹¹ N/C
Explanation:
a)
p(r) = -A exp ( - 2r/a₀)
Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV = -A ₀∫^∞ ₀∫^π ₀∫^2π exp ( - 2r/a₀)r² sinθdrdθd∅
Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e
now using integration by parts;
A = e / πa₀³
p(r) = - (e / πa₀³) exp (-2r/a₀)
Now Net charge inside a sphere of radius a₀ i.e Qnet is;
= e - (e / πa₀³) ₀∫^a₀ ₀∫^π ₀∫^2π r² exp (-2r/a₀)dr
= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b)
Using Gauss's law,
E × 4πa₀ ² = Qnet / ∈₀
E = 4πa₀ ² × Qnet × 1/a₀²
E = 3.466 × 10¹¹ N/C
Answer:
The maximum mass the bar can support without yielding = 32408.26 kg
Explanation:
Yield stress of the material (
) = 200 M Pa
Diameter of the bar = 4.5 cm = 45 mm
We know that yield stress of the bar is given by the formula
Yield Stress = 
⇒
=
---------------- (1)
⇒ Area of the bar (A) =
×
⇒ A =
× 
⇒ A = 1589.625 
Put all the values in equation (1) we get
⇒
= 200 × 1589.625
⇒
= 317925 N
In this bar the
is equal to the weight of the bar.
⇒
=
× g
Where
is the maximum mass the bar can support.
⇒
= 
Put all the values in the above formula we get
⇒
= 
⇒
= 32408.26 Kg
There fore the maximum mass the bar can support without yielding = 32408.26 kg
Answer:
a toy car speed is about 2.5 to 3.5 mph
Answer:
15
Explanation:
P=W/T
T=6sec
W=?
F=60N
S=18m
W=F X S. .s indicate displacement
W=60x18
W=108
So p=108 j/6sec
P=15watt