Answer:
r = 5.08 m
Explanation:
The electric force of attraction or repulsion is given by :
We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.
So, the force from the proton is balanced by the mass of the electron.
r is distance
So, proton have to be at a distance of 5.08 meters above the electron.
Answer:
you could measure several properties of
the unknown liquid and compare them with the properties of known
substances. You might observe and measure such properties as color,
odor, texture, density, boiling point, and freezing point.
The equivalent capacitance between A and B points is 2.5F.
<h3>What is parallel plate capacitor?</h3>
The two parallel plates placed at a distance apart used to store charge when electric supply is on.
The capacitance of a capacitor is given by
C = ε₀ A/d
From the given circuit C1, C2 and C3, C4 are in parallel C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F
C1, C2 = 4 +4 =8F
C3, C4 = 2 +4 =6F
Now , all capacitors are in series.
Total equivalent capacitance is
1 / Ceq = 1/ 8 +1/6 +1/ 9.2
Ceq = 2.5 F
Thus, the equivalent capacitance between A and B points is 2.5F.
Learn more about parallel plate capacitor.
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Answer:
0.35
Explanation:
According to Newton's second law;
\sum Fx = ma
Fm - Ff =ma
Fm is the moving force = Wsin theta
Fm = 4(9.8)sin55
Fm = 32.1N
Ff is the frictional force = nmgcos theta
Ff = n(4)(9.8)cos55
Ff = 22.48n
Acceleration a = 6.0m/s²
Substitute the given values into the formula and get the coefficient of friction
32.11-23.48n = 4(6)
32.11-24= 23.48n
8.11 = 23.48
n = 8.11/23.48
n = 0.35
Hence the coefficient of friction is 0.35
Explanation :
It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.
We know that acceleration is defined as the rate of change of velocity.
Where
dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s
dt is the change in time, dt = 40 s - 30 s = 10 s
So, 
From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. 
.
