Answer:
the weight of the ball is w = 51.94 N ( mass = 5.3 kg)
Explanation:
Following Newton's second law:
net force = mass * acceleration = weight/gravity * acceleration
then denoting 1 and 2 as the first and second lift
F₁ - w= w/g *a₁
F₂ -w = w/g *a₂ = w/g * 2.07a
dividing both equations
(F₂- w)/(F₁ -w)= 2.07
(F₂- w) = 2.07 * (F₁ -w)
1.07*w = 2.07*F₁ - F₂
w = (2.07*F₁ - F₂ )/ 1.07
replacing values
w = (2.07*61.1 N - 70.9 N )/ 1.07 = 51.94 N
then the weight of the ball is w = 51.94 N ( mass = 5.3 kg)
If the potential is given by v = xy - 3z-2, then the electric field has a y-component of X
When the charge is present in any form, a point in space has an electric field that is connected to it. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field.
Each location in space where a charge exists in any form can be considered to have an electric field attached to it. The electric force per unit charge is another name for an electric field. The electric field's equation is given as E = F / Q. Volts per meter (V/m) is the electric field's SI unit. Newton's per coulomb unit is the same as this one.
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Answer: 4nmeter
Explanation: The two observer a and b will measure the same wavelength since the speed of the space craft is very small compared with the speed of light c. That is
V which is the speed of space craft 15000km/s = 15000000m/s
Comparing this with the speed of light c 3*EXP(8)m/s we have
15000000/300000000
= 0.05=0.1
Therefore the speed of the space craft V in terms of the speed of light c is 0.1c special relativity does not apply to object moving at such speed. So the wavelength would not be contracted it will remain same for both observers.
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
