Answer:
-0.64525g
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 95 km/h
v = Final velocity = 0 km/h
s = Displacement
a = Acceleration
Equation of motion
Converting to m/s²
g = Acceleration due to gravity = 9.81 m/s²
Dividing both the accelerations, we get
Hence, acceleration of the car is -0.64525g
Answer:
B. stearothermophilus and S. ruber
Explanation:
B. stearothermophilus and S. ruber
In solar evaporation ponds the temperature is higher and the salt concentration is also higher because of the water evaporated so sunder such extreme conditions this hybrid bacteria is capable of surviving. B. stearothermophilus is thermophilus bacteria which grows at high temperature and S. ruber is halophilic bacteria which grows in saline environment. So, these two bacteria best suited for the above hybrid condition.
Series Circuit
A series circuit there is only one path for the electrons to flow (see image of series circuit). The main disadvantage of a series circuit is that if there is a break in the circuit the entire circuit is open and no current will flow. An example of a series would the the lights on many inexpensive Christmas trees. If one light goes out all of them will.
Parallel Circuit
In a parallel circuit the different parts of the electric circuit are on several different branches. There are several different paths that electrons can flow. If there is a break in one branch of the circuit electrons can still flow in other branches (see image of parallel circuit). Your home is wired in a parallel circuit so if one light bulb goes out the other will stay on.
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Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
