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VARVARA [1.3K]
3 years ago
8

A hotel wants to put a fence around a circular spa. The radius of the spa is 5/√2-1 feet. If the fence is built along the immedi

ate edge of the spa, what is the perimeter of the fence? (Recall that the
perimeter of a circle is 2πr , where r is the radius of the circle.)
Physics
1 answer:
Vlad [161]3 years ago
5 0
It is important to look at all the information's that are given in the question very closely. Let us write them write first.

Radius of the spa = <span>5/√2-1 feet
Now
Perimeter of the circle = </span><span>2πr
                                    = 2</span>π (5/√2-1) 
                                    = <span>π(5/(√2-1)*(√2+1)/(√2+1) </span>
<span>                                    = 2π(5/(√2+1))/(2-1) </span>
<span>                                    = 10π(√2+1)  

I hope that the procedure is clear enough for you to understand.</span>
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lesya692 [45]

Answer:

ozone, letter b

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3 years ago
Find the total power developed in the circuit. Express your answer using three significant figures and include the appropriate u
BabaBlast [244]

Answer:

37 W

Explanation:

Power is the time rate of dissipation or absorbing energy. The power supplied or absorbed by an element is the product of the current flowing through the element and the voltage across the element. Power is measured in watts. If the power is positive then it is absorbed by the element and if it is negative then it is supplied by the element.

Power = voltage * current

For element A: Power = 36 V * -4 A = -144 W

For element B: Power = -20 V * -4 A = 80 W

For element C: Power = -24 V * 4 A = -94 W

For element D: Power = -80 V * -1.5 A = 120 W

For element E: Power = 30 V * 2.5 A = 75 W

The total power developed in the circuit = sum of power through the element = (-144 W) + 80 W + (-94 W) + 120 W + 75 W = 37 W

6 0
3 years ago
Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.
Nana76 [90]

Answer:

\sigma_i=1.06*10^{-6}C

Explanation:

When the space is filled with dielectric, an induced opposite sign charge appears on each surface of the dielectric. This induced charge has a charge density related to the charge density on the electrodes as follows:

\sigma_i=\sigma(1-\frac{E}{E_0})

Where E is the eletric field with dielectric and E_0 is the electric filed without it. Recall that \sigma is given by:

\sigma=\epsilon_0E_0

Replacing this and solving:

\sigma_i=\epsilon_0E_0(1-\frac{E}{E_0})\\\sigma_i=8.85*10^{-12}\frac{C^2}{N\cdot m^2}*3.40*10^5\frac{V}{m}*(1-\frac{2.20*10^5\frac{V}{m}}{3.40*10^5\frac{V}{m}})\\\sigma_i=1.06*10^{-6}C

3 0
3 years ago
Two objects collide, one traveling downwards with a mass of 2.0 kg, and the other traveling upwards with a mass of 1.0 kg. Both
vladimir1956 [14]

Answer:

-1 m/s

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m_d v_d+m_u v_u=(m_d + m_u)v_c

Where subscripts d and u represent downwards and upwards respectively while c is common hence common velocity which is unknown in this case. Taking downwards as negative while upwards as positive then substituting -3 m/s for downward speed while 3 m/s for upward speed, the masses as given then

(2*-3)+(1*3)=(2+1)v

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4 0
3 years ago
How do you calculate the speed of a 3.1eV photon and a 3.1eV electron?
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Explanation:

We need to find the speed of a 3.1eV proton and a 3.1eV electron.

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For electron,

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v_e=\sqrt{\dfrac{2\times 3.1\times 1.6\times 10^{-19}\ J}{9.1\times 10^{-31}\ kg}}\\\\v_e=1.44\times 10^6\ m/s

Hence, this is the required solution.

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3 years ago
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