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Bas_tet [7]
3 years ago
13

A bug zapper consists of two metal plates connected to a high-voltage power supply. The voltage between the plates is set to giv

e an electric field slightly less than 1×10⁶ V/m . When a bug flies between the two plates, it increases the field enough to initiate a spark that incinerates the bug.
If a bug zapper has a 7000 V power supply, what is the approximate separation between the plates?
(a) 7×10⁻² cm
(b) 0.7 cm
(c) 7 cm
(d) 70 cm
Physics
1 answer:
Allisa [31]3 years ago
4 0

Answer:

(b) 0.7 cm

Explanation:

E = Electric field = 1\times 10^6\ V/m

V = Voltage = 7000 V

d = Seperation between the plates

Electric field is given by

E=\dfrac{V}{d}

\\\Rightarrow d=\dfrac{V}{E}\\\Rightarrow d=\dfrac{7000}{1\times 10^6}\\\Rightarrow d=0.007\ m=0.7\ cm

The approximate separation between the plates is 0.7 cm

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Mathematically, relation between force, area and pressure is given by...
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Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range
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Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

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3 years ago
microwave ovens rotate at a rate of about 6.3 rev/min. what is this in revolutions per second? (you do not need to enter any uni
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\omega = 6.3 \times \frac{1 \: rev}{60 \: seconds}

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7 0
1 year ago
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
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Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet x=\frac{4}{12} =\frac{1}{3}feet

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

W=kx ⇒ k=\frac{W}{x}

The spring constant , k=\frac{24}{1/3}

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    m\frac{d^2x}{dt} +kx=0

    \frac{3}{4} \frac{d^2x}{dt} +72x=0

  \frac{d^2x}{dt} +96x=0

Auxiliary equation is, m^2+96=0

                                 m=\sqrt{-96}

                               =\frac{+}{} i4\sqrt{6}

Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

                                 x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2  4\sqrt{6} cos4\sqrt{6t}

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                    =-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2 4\sqrt{6} cos4\sqrt{6(0)} =0

                                                    c_2 4\sqrt{6} =0

                                     c_2=0

Therefore , x(t)=c_1 cos 4\sqrt{6t}

Since , the mass is released from the rest from 4 inches

                    x(0)= -4 inches

c_1 cos 4\sqrt{6(0)} =-\frac{4}{12} feet

   c_1=-\frac{1}{3} feet

Therefore , the equation of motion is  -\frac{1}{3} cos4\sqrt{6t}

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