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ddd [48]
3 years ago
15

ruto is 12 years old.in three years time he will be ⅓ of his father present age.how old was his father 12 years ago​

Physics
1 answer:
max2010maxim [7]3 years ago
4 0

Answer: 33

Explanation:

12+3=15, so in 3 years time Ruto will be 15.

Because 15 is said to be 1/3 times his fathers present age (x), you can plug that into an equation.

15=1/3(x)

Then solve the equation.

(3)15=[1/3(x)]•3

45=x

His father is currently 45, so if you subtract 12 from that, you get 33.

Ruto’s father was 33 years old 12 years ago.

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What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x
lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, U_{electric,b} - U_{electric,a} as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2×10^{-6} J

(b) KE = 2×10^{-6} J

Explanation: <u>Potential</u> <u>Energy</u> (U) is the amount of work done due to its position or condition and its unit is Joule (J). <u>Kinetic</u> <u>Energy</u> (KE) is the ability to do work by virtue of velocity and the unit is also (J). <u>Mechanical</u> <u>Energy</u> is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2×10^{3} N/C

q = 2×10^{-9} C

d = 50 - (-30) = 80×10^{-2} = 8×10^{-1}m

ΔU = U_{electric,b} - U_{electric,a} = Eqd

U_{electric,b} - U_{electric,a} = 2×10^{3} .  2×10^{-9} . 8×10^{-1}

ΔU = 3.2×10^{-6} J

(b) Mechanical Energy is constant, so:

KE_{i} + U_{i} = KE_{f} + U_{f}

Since the initial position is zero and there is no initial kinetic energy:

KE_{f} = - U{f}

KE_{f} = - (2×10^{3}. 2×10^{-9} . 5×10^{-1})

KE_{f} = - 2.10^{-6} J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

8 0
3 years ago
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