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umka21 [38]
3 years ago
11

Hey anyone here I'm ban 2 mint ago my I'd is aron7 I talk to Rachel anybody present therw​

Physics
2 answers:
pishuonlain [190]3 years ago
8 0

sign out and log in again...if does not work then make a new account

gtnhenbr [62]3 years ago
5 0

Answer:

follow me and mark me brainliest

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Someone please help me with these questions! (The ones in the picture) Please I am super confused!
lozanna [386]

Answer:

c-d

Explanation:

3 0
2 years ago
A painter stands a horizontal platform which has a mass of 20kg and is 5m long, the platform is suspended by two vertical ropes
Lerok [7]

Answer:

R = 715.4 N

L =  166.6 N

Explanation:

ASSUME the painter is standing right of center

Let L be the left rope tension

Let R be the right rope tension

Sum moments about the left end to zero. Assume CCW moment is positive

R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0

R = 715.4 N

Sum moments about the right end to zero

20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0

L =  166.6 N

We can verify by summing vertical forces

116.6 + 715.4 - (70 + 20)(9.8) ?=? 0

                                             0 = 0  checks

If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.

5 0
3 years ago
What is one use for infrared waves? A radar B Medical imaging C Thermal imaging cameras
Vilka [71]

Answer:

Im answering for free points sry

Explanation:

...

7 0
2 years ago
A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.
noname [10]

Answer:

The value is  \rho_s  =  4.026 *10^{-6} \  C/m^2

Explanation:

From the question we are told that

   The radius of the inner conductor  is  r_1 = 1 \ mm =  0.001 \ m

    The radius of the outer conductor is  r_2 = 3 \ mm = 0.003 \  m

    The potential at the outer conductor is  V = 1.5 kV  =  1.5 *10^{3} \  V

Generally the capacitance per length of the capacitor like set up of the two conductors is

      C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}

Here \epsilon_o is the permitivity of free space with value  \epsilon_o =  8.85*10^{-12} C/(V \cdot m)

=>   C= \frac{2 *  3.142  * 8.85*10^{-12}  }{ ln [\frac{0.003}{0.001} ]}

=>   C= 50.6 *10^{-12} \  F/m

Generally given that the potential  of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line  charge density of the outer  conductor is mathematically represented as

      \rho_l  =  C *  V

=>   \rho_d  =  50.6*10^{-12} *  1.5*10^{3}

=>   \rho_d  =  7.59*10^{-8} \  C/m

Generally the surface charge density is mathematically represented as

        \rho_s  =  \frac{\rho_l }{2 \pi * r_2 }    here 2 \pi r = (circumference \ of \ outer \  conductor  )

=>    \rho_s  =  \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }

=>    \rho_s  =  4.026 *10^{-6} \  C/m^2

3 0
2 years ago
3. Why is static electricity not useful as a power
shtirl [24]
The answer is B, because all energy is released at once in static electricity.

There’s a quizlet that mentions these questions, if you are having trouble. I’d suggest to give them a look.
3 0
2 years ago
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