If an electron is accelerated through a potential difference of 45.5 volts, then what is the velocity acquired by it?
Whenever a charged particle q is accelerated through a potential difference of V, its kinetic energy increases by an amount of qV. This you can deduce simply by applying the total mechanical energy conservation principle on the system as follows:Now in the given problem q = e = charge of an electron, m = mass of an electron and potential difference V = 45.5 volts. On putting these values in the above expression we getv = 4 × 10^6 m/s.Regards!
Two bodies of masses m and 4m are moving at equal moments. What is the ratio of their kinetic energy?
Lets say the velocity of ‘m’ is ‘v’ and velocoty of ‘4m’ is ‘u’Therefore, mv = 4mu ———————— GivenSo, v = 4uNow we know that kienetic energy = (1/2) mv^2Thus for the ratio, we need to calculate(1/2 * m * (4u)^2) / (1/2 * 4m * u^2)Turns out to be 4/1.Which means the ratio of keinetic energies of both masses is 4:1
A hole is made at the bottom of the tank filled with water. if the total pressure at the bottom is 3 atmospheres, then what is the velocity of efflux?
you can apply Bernoulli’s equation to get the velocity. the dynamic pressure is 2 atmosphere. sov=sqrt(2*2*101325/1000)=20.13 m/sec
There are four equal charges that are kept at each corner of a square. How do I calculate the force experienced by one charge? What is the final expression?
By Coulombs law, equal force is acting on each these four charges. Because of which it will always form a square with four charges at the corners of the square.Force acting on charge is 3Kq[math]2/r2[/math]*2 is squareThis can be demonstrated. Three charges for a triangle etc.
8 spherical rain drops of equal size are falling vertically through air with uniform speed of 1m/s. What would be the uniform speed if these drops were to combine to form one large spherical drop?
Falling with uniform speed means the drops have achieved terminal speed.Let us find the radius [math]R[/math] of the bigger drop in terms of the radius [math]r[/math] of the smaller drop. We use conservation of volume to arrive at the result.[math]8.\frac 43 \pi r^3 = \frac 43 \pi R^3 \Rightarrow R = 2r[/math]Now terminal speed is directly proportional to the square of the radius. Thus[math]\frac{v_1}{v_2} = \frac{r^2}{R^2}=\frac{r^2}{(2r)^2} \Rightarrow v_2 = 4v_1[/math]