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agasfer [191]
3 years ago
15

Refer to the picture of the nutcrackers. The small nut is placed close to the pivot of the nutcracker and the larger nut is plac

ed farther from the pivot. Comparing the nutcracker cracking the larger nut to the smaller nut, describe the difference in your hand's force and the distance from the nut between the two.
A. more force required, less distance moved

B. less force, less distance

C. more force, more distance

D. less force, more distance
Physics
2 answers:
Pachacha [2.7K]3 years ago
7 0
D.
Due to the greater distance from the pivot, less force is required to crack the nut as
Moment = Force x distance.
Neko [114]3 years ago
5 0
<h2><u><em>A. more force required, less distance moved</em></u> </h2>

i just took the test and i put the answer that the other person put and it was wrong

*would've gotten a 100% if i didnt get this question wrong...*

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A charge of 63.0 nC is located at a distance of 3.40 cm from a charge of -47.0 nC. What are the x- and y-components of the elect
Serjik [45]

Answer:

Ep_x = 288.97*10^3\frac{N}{C}

Ep_y = 2770.6*10^3\frac{N}{C}

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/r²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

r: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.

Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.

Known data

q₁ = 63 nC = 63×10⁻⁹ C

q₂ = -47 nC = -47×10⁻⁹ C

k = 8.99*10⁹ N×m²/C²

d₁ = 1.4cm = 1.4×10⁻² m

d₂ = 3.4cm = 3.4×10⁻² m

Calculation of r and β

r=\sqrt{d_1^2 + d_2^2} = \sqrt{(1.4*10^{-2})^2 + (3.4*10^{-2})^2} = 3.677*10^{-2}m

\beta = tan^{-1}(\frac{d_1}{d_2}) = tan^{-1}(\frac{1.4}{3.4}) = 22.38^o

Problem development

Ep: Total field at point P due to charges q₁ and q₂.

Ep = Ep_x i + Ep_y j

Ep₁ₓ = 0

Ep_{2x}=\frac{-k*q_2*Cos\beta}{r^2}=\frac{8.99*10^9*47*10^{-9}*Cos(22.38)}{(3.677*10^{-2})^2}=288.97*10^3\frac{N}{C}

Ep_{1y}=\frac{-k*q_1}{d_1^2}=\frac{8.99*10^9*63*10^{-9}}{(1.4*10^{-2})^2}=2889.6*10^3\frac{N}{C}

Ep_{2y}=\frac{-k*q_2*Sen\beta}{r^2}=\frac{-8.99*10^9*47*10^{-9}*Sen(22.38)}{(3.677*10^{-2})^2}=-119*10^3\frac{N}{C}

Calculation of the electric field components at point P

Ep_x = Ep_{1x} + Ep_{2x} = 0 + 288.97*10^3 = 288.97*10^3\frac{N}{C}

Ep_y = Ep_{1y} + Ep_{2y} = 2889.6*10^3 - 119*10^3 = 2770.6*10^3\frac{N}{C}

6 0
3 years ago
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