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likoan [24]
3 years ago
13

How to find the angle marked

Physics
1 answer:
Sedbober [7]3 years ago
4 0
#6).  The tangent of an angle in a right triangle is

                                           (opposite side) / (adjacent side) .

The side opposite the marked angle is  4cm.
The side adjacent to the marked angle is  12cm.

The tangent of the marked angle is   (4cm / 12cm)  =  1/3 .

The marked angle is the angle whose tangent is 1/3 .
You can find that online, or with your calculator, slide rule, Curta, or book.


#7).  The cosine of an angle in a right triangle is

                                           (adjacent side) / (hypotenuse) .

The side adjacent to the marked angle is  15m.
The hypotenuse is  19m.

The cosine of the marked angle is   (15m / 19m)  =  about 0.78947...

The marked angle is the angle whose cosine is 0.78947... .
You can find that online, or with your calculator, slide rule, Curta, or book.
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Due to an influx of wild animals, you need to enclose your garden. You have 500 feet of fencing and will enclose the garden on t
elena55 [62]

Answer:

so  length of the fence  is 250 feet

Explanation:

Given data

fencing (L) = 500 feet

enclose the garden = 3 sides

to find out

length of the fence

solution

we consider barn length x and width y

so we say  x +2y = 500   ........1

we want to maximize length i.e xy = f

so we say

df/dx = ƛ × dL/dx

y =  ƛ × 1

and

df/dy = ƛ × df/dy

x = ƛ × 2

from equation 1 we can say

2 ƛ + 2 ƛ = 500

so  ƛ = 500/4 = 125

so that x = 2× 125 ,  y = 125

so x = 250 , y = 125

so  length of the fence  is 250 feet

3 0
4 years ago
n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble
Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm  is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is    U_f -U_i = 16 *10^{-7} J

Explanation:

From the question we are told that

     The radius of the soap bubble  is  R_o =  7.45 \ cm =  \frac{7.45}{100} =  0.0745 \ m

      The potential of the soap bubble is  V_1  =307.0 V

      The new radius of the soap bubble  is R_1 =  0.19 * 7.45=1.4155\ cm = 0.014155 \ m

The initial electric potential is mathematically represented as

     U_i  = \frac{V_1^2 R_o }{2k }

The final  electric potential is mathematically represented as

    U_f  = \frac{V_2^2 R_1 }{2k }

The initial potential is mathematically represented as

     V_1 =  \frac{kQ}{R_o}

The final  potential is mathematically represented as

        V_2 =  \frac{kQ}{R_1}

Now  

         \frac{V_2}{V_1}  =  \frac{R_o}{R_1}

substituting values

        \frac{V_2}{V_1}  =  \frac{7.45}{1.4155} =   \frac{1}{0.19}

=>      V_2 =  \frac{V_1}{0.19}

    So

         U_f  = \frac{V_1^2 R_2 }{2k * 0.19^2}

Therefore

        U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }

       U_f -U_i =     \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]

where k is the coulomb's constant with value 9*10^{9} \  kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

       U_f -U_i =     \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]

       U_f -U_i = 16 *10^{-7} J

           

     

8 0
3 years ago
How far apart must two protons be if the electrical force acting on either one is equal to its weight on the earth's surface whe
BARSIC [14]

Answer:

G M m / R^2 = F   force of attraction for proton

K e^2 / r^2 = F     force of attraction between 2 protons

G M m / R^2 = K e^2 / r^2

r^2 = K e^2 R^2 / (G M m)

r^2 = 9E9 * (1.6E-19)^2 * (6.37E6)^2 / (6.67E-11 * 5.98E24 * 1.7E-27)

r^2 = 9 * 2.56 * 40.6 / (6.67 * 5.98 * 1.7) * 10^-19

r^2 = 1.38 * 10^-30

r = 1.17E-15 m

7 0
3 years ago
Is magnesium a metal ?
tatuchka [14]
Yes Magnesium Its Metal 
8 0
3 years ago
A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitud
polet [3.4K]

Answer:

The net charge on the shell is 30x10^-9C

Explanation:

Pls see attached file

8 0
4 years ago
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