in this since your volume remains at a constant you'll need to use Gay-Lussacs law, p1/t1=p2/t2.
your temp should be converted in kelvin
variables:
p1=3.0×10^6 n/m^2
t1= 270k
just add 273 to your celcius
p2= ? your solving for this
t2= 315k
then you set up the equation
(3.0×10^6)/270= (x)(315)
you then cross multiply
(3.0×10^6)315=270x
distribute the 315 to the pressure.
9.45×10^8=270x then you divide 270 o both sides to get
answer
3.5×10^6 n/m^2
The image as shown here can here can be used to describe charging by induction.
<h3>What is a charge?</h3>
A charge may be positive or negative. One of the methods of transferring a charge is by induction.
In this case, an objects induces an opposite charge on a material. The image as shown here can here can be used to describe charging by induction.
Learn more about charging by induction:brainly.com/question/10254645
#SPJ4
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
<span>To answer the question above, if the day sky is clear it collects radiation. If air is dry snow sublimates faster. If both cases overlap it disappears faster where ever coldest. Thank you for posting your question here. I hope my answer helps. </span>
Answer:
Approximately 
(given that the magnitude of this charge is 
.)
Explanation:
If a charge of magnitude 
is placed in an electric field of magnitude 
, the magnitude of the electrostatic force on that charge would be 
.
The magnitude of this charge is 
. Apply the unit conversion 
:
.
An electric field of magnitude 
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
