Answer:
Explanation:
Initial separation of plate = d
final separation = 2d
The capacitance of the capacitor will reduce from C to C/2 because
capacitance = ε A / d
d is distance between plates.
As the batteries are disconnected , charge on the capacitor becomes fixed .
Initial charge on the capacitor
= Capacitance x potential difference
Q = C ΔV
Final charge will remain unchanged
Final charge = C ΔV
Final capacitance = C/2
Final potential difference = charge / capacitance
= C ΔV / C/2
= 2 ΔV
Potential difference is doubled after the pates are further separated.
Answer:
she should add solute to the solvent
Explanation:
Given data :
Mass of the sodium chloride, = 20.0 g
Concentration of the solution = 10 g/L
Volume of 20.0 g of sodium chloride = 7.50 mL
Now, from the concentration, we can conclude that for 10 g of sodium chloride volume of the solution is 1 L
thus, for 20 g of sodium chloride volume of the solution is 2 L or 2000 mL
also,
Volume of solution = Volume of solute(sodium chloride) + volume of solvent (water)
thus,
2000 mL = 7.5 mL + volume of solvent (water)
or
volume of water = (2000 - 7.5) mL
or
volume of water = 1992.5 mL
or
volume of water = 199.25 L ≈ 199 L
Answer:
no I don’t think there can be so my answer is No.
Okay then yes sorry that I must have gotten it wrong before.
Explanation:
347÷134=2.589552239 meters per second
2.589552239×60= 155.3731343 meters per hour
155.3731343 meters per hour= 0.096544389701642 miles per hour
hopefully this was right.
(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>

Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

<h3>Resultant speed of the ball and crosswind</h3>

<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671