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ASHA 777 [7]
3 years ago
10

Consider what will happen when a bar magnet is pushed toward the coil. when the coil "feels" the changing magnetic field caused

by the approach of the bar magnet, it will thereby become energized; a current will begin to flow through it, and the ammeter will register that current. this current is called an induced current because it has been induced by the person
Physics
1 answer:
notka56 [123]3 years ago
4 0
The question just basically explained what happens
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A ski jumper travels down a slope and
AleksandrR [38]

Answer:

304.86 metres

Explanation:

The x and y cordinates are dcos\theta and dsin\theta respectively

The horizontal distance travelled, x=v_{ox}t=dcos\theta

Making t the subject, t=\frac{dcos\theta}{v_{ox}}

Since y=0.5gt^2=dsin\theta, we substitute t with the above and obtain

0.5g(\frac{dcos\theta}{v_{ox}})^2=dsin\theta

Making d the subject we obtain

d=\frac{2v_{ox}^2sin\theta}{gcos^2\theta}

d=\frac{2*30^2sin48}{9.8cos^248}

d=304.8584

d=304.86m

5 0
3 years ago
What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in 60 min? assume first-order kin
LenKa [72]
The amount left of a given substance can be calculated through the equation,
                                              A = (A0) x 0.5^n/h
From the given scenario, 
                                           A/A0 = 0.75 = 0.5*(60/h)
The value of h from the equation is 144.565 minutes. 
6 0
4 years ago
Give examples for the following (You may do as many as you want):
valina [46]

Answer:

Explanation:6

6 0
3 years ago
A temperature of 200°F is equivalent to approximately
aev [14]
<span>93.3°C
A temperature in Fahrenheit (°F) can be converted to Celsius (°C), using the formula
[°C] = ([°F] − 32) ×  5⁄9. Here we have to convert a temperature of 200°F in to Celsius. Thus Subtract 32 from Fahrenheit and multiply by 5 then divide by 9 . That is (200°F - 32) × 5/9=168 × 5/9
                                          =840/9
                                          =93.333333333°C
                                          = 93.3°C</span>
7 0
3 years ago
Read 2 more answers
A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius
chubhunter [2.5K]

Answer:

a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2

b) See the picture

Explanation:

We can use Gauss's law to find the electric field in all the regions:

EA = qen/e0 where qen is the enclosed charge

Remember that the electric field everywhere outside a sphere is:

E(r) = q/(4*pi*eo*r^2) = Kq/r^2

a)

  1. For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0                       EA = 0/e0 = 0                                                                                                    E = 0
  2. For R < r < 2R: Here the enclosed charge is equal Q                                      E =  Q/(4*pi*eo*r^2) = KQ/r^2      
  3. For r > 2R: Here the enclosed charge is equal 2Q                                              E =  Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2

b)  At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance

7 0
3 years ago
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