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ASHA 777 [7]
3 years ago
13

A step down transformer has a primary coil consisting of 700 tons and a secondary coil consisting of 300 times. What is the pote

ntial difference in the second circuit, if the primary coil is supplied with an effective AC potential difference of 110 V?
Physics
1 answer:
Pie3 years ago
6 0

Answer is 47.14 v

by the relation

\frac{V out}{V in } = \frac{N secondary}{N primary }

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If i punch my self and it will hurt am i weak or strong
Natasha2012 [34]

Answer:

You're strong.

Explanation:

I've been thinking of this for quite a while, and I realized that your body has a certain limit to how much pain it can take. So, punching yourself extremely hard will cause pain, because that's your body's reaction to immense pressure being put on it. But, the fact that you punched yourself so hard that it hurts, shows that you are capable of applying so much pressure; therefore, you are strong.

6 0
3 years ago
A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.
Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

a=\frac{v^{2} }{r}, where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

a=\frac{3.5^{2} }{8.5} \\a=1.4

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

7 0
3 years ago
A plane is going at a speed of 300 km/h at 63 W of N. The wind hits the plane at a direction of 65 km/h at 52 S of E. What is th
katovenus [111]
It’s around the g force so it’s gonna be around 54 km/h
3 0
2 years ago
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
Afootball of mass 550g is at rest on the ground the football is kicked with a force of 108 newton the footballers boot is in con
d1i1m1o1n [39]

If "0.3 minute" is correct, then it's 9,543,272 Joules.

If it's supposed to say "0.3 SECOND", then the KE is 2,651 Joules.

6 0
3 years ago
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