Answer:
Work and Kinetic Energy
A B
3. A 0.180 kg balls falls 2.5 m. How much work does the force of gravity do on the ball? 4.41 J
4. A forklift raises a box 1.2 m doing 7.0 kJ of work on it. What is the mass of the box? 595.24 kg
5. How much work does the force of gravity do when a 25 N object falls a distance of 3.5 m? 87.5 J
Explanation:
Responda:
1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5
Explicação:
Dado o seguinte:
Carga (q) = 3uC = 3 × 10 ^ -6C
Força elétrica (Fe) = 18N
Intensidade do campo elétrico (E) =?
1)
Lembre-se:
Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)
Fe = qE; E = Fe / q
E = 18N / (3 × 10 ^ -6C)
E = 6N / 10 ^ -6C
E = 6 × 10 ^ 6NC ^ -1
2)
Lembre-se:
E = kQ / r ^ 2
E = intensidade do campo elétrico
Q = carga de origem
r = distância de espera = 30cm = 30/100 = 0,3m
K = 9,0 × 10 ^ 9
6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2
9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09
Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9
Q = 0,06 × 10 ^ (6-9)
Q = 0,06 × 10 ^ -3
Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC
It's total kinetic energy
Answer:
The distance is 
Explanation:
From the question we are told that
The wavelength of the light is 
The distance between the slit is 
The between the first and second dark fringes is 
Generally fringe width is mathematically represented as
Where D is the distance of the slit to the screen
Hence
substituting values
Show me the graphs and i would be glad to help u
