Answer:
2.92 x 10¹² electrons
Explanation:
given,
distance between two plastic ball, r = 17 cm
r = 0.17 m
Force of attraction = F = 68 mN
F = 0.068 N
number of electron transferred from one ball to another.
using Coulomb Force equation
q² = 2.1835 x 10⁻¹³
q = 4.67 x 10⁻⁷ C
now, number of electron
e is the charge of electron which is equal to 1.6 x 10⁻¹⁹ C
N = 2.92 x 10¹² electrons
electrons were transferred from one ball to the other is 2.92 x 10¹²
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s
a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49
Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.
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This is incomplete question Complete Question is:
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?
Solution :
The relationship between the strength of magnetic field and the radiusof a charged particle's path is obtained through Newton's second law, which is given by :
F = ma
F = qvB and 
Substituting these values in the second law of Newton,
Now solving for B, we get:
= 0.261 T
The field strength can be obtained by using the technology of today.
Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:
simplify by 2
Answer:
B.) 1/2 cm
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I Hope this helps, greetings ... Dexteright02! =)
Answer:
seconds
Explanation:
T = 1 / f
where T is the period and f is frequency.
Here the frequency is 512 Hz.
<em>Using the formula:</em>
T = seconds
