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Degger [83]
3 years ago
11

4. Three methods that people use are:

Physics
1 answer:
Mila [183]3 years ago
6 0
The answers to question 4 d
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how much work is done in moving a charge of 3 coulomb for a point at the volts 115 to a point at 125 volts​
Alona [7]

Answer:

We use the Formula,

Work done = total chargr×potential difference

W=Q(total)× V

Here

Q= total charge= Give 3c

V=potential difference= 10v

so the work done in moving the charge from A to B will be,

w=3×10

w= 30 joules

8 0
3 years ago
Gamma rays are dangerous because they
lina2011 [118]

Answer:

A.

Explanation:

A large fraction of astronomical gamma rays are screened by Earth's atmosphere. Gamma rays are ionizing radiation and are thus biologically hazardous. Due to their high penetration power, they can damage bone marrow and internal organs.

The extremely high energy of gamma rays allows them to penetrate just about anything. They can even pass through bones and teeth. This makes gamma rays very dangerous. They can destroy living cells, produce gene mutations, and cause cancer.

8 0
3 years ago
Define the term unit sector?​
Inessa05 [86]

a single two-dimensional cross-sectional imagine made using real-time technology

8 0
3 years ago
At 15°C air is transmitted <br>at 340 m/s. Express this speed<br>in Kilometers per hour.​
EleoNora [17]

Answer:

1224km/hr

Explanation:

To convert from m/s to km/hr

1000m = 1km

Divide both sides by 1000

1m = 1/1000 km................. (1)

60×60 seconds = 1 hr

3600s = 1hr

Divide both sides by 3600

1s = 1/3600 .............(2)

Divide (2) by (1)

1m/s =  1/1000 ÷ 1/3600 km/hr

1m/s = 1/1000 × 3600/1  km/hr

1m/s = 3600/1000  km/hr

1m/s = 3.6 km/hr .............(3)

To convert 340m/s to km/hr

Multiply (3) by 340

1× 340m/s = 3.6 × 340 km/hr

340m/s = 1224km/hr

I hope this was helpful, please mark as brainliest

7 0
4 years ago
Read 2 more answers
A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit
djverab [1.8K]

Answer:

The current is  I  = 6.68 \  A

Explanation:

From the question we are told that  

     The radius of the loop is  r =  6 \ cm  = 0.06 \ m

     The  earth's magnetic field is B_e =  0.7G=  0.7  G * \frac{1*10^{-4} T}{1 G}  = 0.7 *10^{-4} T

      The  number of turns is  N  =1

Generally the magnetic field generated by the current in the loop is mathematically represented as

        B  =  \frac{\mu_o  * N  *  I}{2 r }

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         B  =  B_e

=>     B_e =  \frac{\mu_o  *  N  *  I  }{ 2 * r}

     Where  \mu is the permeability of free space with value  \mu _o  =   4\pi * 10^{-7} N/A^2

       0.7  *10^{-4}=  \frac{ 4\pi * 10^{-7}  * 1 * I}{2 * 0.06}

=>     I  =  \frac{2 *  0.06 *  0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}

       I  = 6.68 \  A

3 0
4 years ago
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