Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
Answer:
Stephen William Hawking
Explanation:
Stephen William Hawking was a popular theoretical physicist born on 8th January 1942.
He served as the professor of mathematics at the Cambridge university
He was director at the center for theoretical cosmology.
He has theorized that the black holes will emit radiations and the radiations is known by his name Hawking's radiation.
His book 'The Brief History of Time' appeared as one of the best seller according to the Sunday Times.
Hawking's was diagnosed for a rare disease named as motor neuron disease.
He received numerous awards and elected as the fellow of royal society.
He passed away on 14th March 2018.
C. Oxygen has a different attraction for electrons in H2O than in CO2
the covalent bond between Hydrogen-Oxygen and Carbon-Oxygen is different because the difference of electronegativity (the tendency of an atom to attract a shared pair of electrons towards itself) between Hydrogen and Oxygen is a lot higher than the difference of electronegativity between Carbon and Oxygen. This difference in electronegativity results in separation of partial electric charge creating polar bonds and giving the molecule a polarity.
Answer:
(a) a= 0.139 m/s²
(b) d= 4.45 m
(c) vf= 1.1 m/s
Explanation:
a) We apply Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass (kg)
a : acceleration (m/s²)
Data
F₁= +2.05 * 10³ N : forward push by a motor
F₂= -1.87* 10³ N : resistive force due to the water.
m= 1300 kg
Calculation of the acceleration of the boat
We replace data in the formula (1):
∑F = m*a
F₁+F₂= m*a
a= 0.139 m/s²
b) Kinematics of the boat
Because the boat moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+at Formula (3)
Where:
d:displacement in meters (m)
t : time interval (s)
v₀: initial speed (m/s)
vf: final speed (m/s)
a: acceleration (m/s²
)
Data
v₀ = 0
a= 0.139 m/s²
t = 8 s
Calculation of the distance traveled by the boat in 8 s
We replace data in the formula (2)
d= v₀t+ (1/2)*a*t²
d= 0+ (1/2)*(0.139)*(8)²
d= 4.45 m
c) Calculation of the speed of the boat in 8 s
We replace data in the formula (3):
vf= v₀+at
vf= 0+( 0.139)*(8)
vf= 1.1 m/s
Answer:
The answer to your question is:
a) t1 = 2.99 s ≈ 3 s
b) vf = 39.43 m/s
Explanation:
Data
vo = 10 m/s
h = 74 m
g = 9.81 m/s
t = ? time to reach the ground
vf = ? final speed
a) h = vot + (1/2)gt²
74 = 10t + (1/2)9.81t²
4.9t² + 10t -74 = 0 solve by using quadratic formula
t = (-b ± √ (b² -4ac) / 2a
t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)
t = (-10 ± √ 1550.4 ) / 9.81
t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81
t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81
t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are
no negative times.
b) Formula vf = vo + gt
vf = 10 + (9.81)(3)
vf = 10 + 29.43
vf = 39.43 m/s
