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o-na [289]
4 years ago
13

How do you find input energy with only the efficiency and output energy given?

Physics
1 answer:
iris [78.8K]4 years ago
8 0

Answer:

power output/transformer efficiency multiply by 100%

Explanation:

derrive the formula from efficiency=power output/power input multiply by 100%

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The __________ are tubular, fluid-filled structures that provide information about head rotations and movements.
shusha [124]

Answer: C

Explanation:

The semi-circular canals are tubular, fluid-filled structures that provide information about head rotations and movements.

The semi-circular canals match the description. Therefore, it is the answer.

Hope this helps!

4 0
3 years ago
Read 2 more answers
A projectile is launched in the horizontal direction. It travels 2.050 m horizontally while it falls 0.450 m vertically, and it
kari74 [83]

Answer:

0.303s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

4 0
4 years ago
a bungee jumper momentarily comes to rest at the bottom of the dive before he springs back upward. at that moment, is the bungee
barxatty [35]

Answer:

Explanation:

No, the bungee jumper is not at equilibrium.

This can be explained when we consider a bungee jumper as a mass that is undergoing simple harmonic motion. At extreme points i.e. at the bottom, the velocity of the jumper is zero but not the acceleration because it is acting in the opposite direction that is why the jumper moves upward.

3 0
4 years ago
Read 2 more answers
a 10. kg ball thrown into the air. it is going 3.0 m/s when thrown. how much potential energy will it have at the top?
Sever21 [200]

Answer:

45 j

Explanation:

At the top, the KE will have been converted to PE

KE = 1/2 mv^2 =PEat-top = 1/2 10 *3^9 = 45 j

6 0
2 years ago
Calculate the nuclear binding energy per nucleon for 136^Ba if its nuclear mass is 135.905 amu.
kompoz [17]

Answer:

1.312 x 10⁻¹² J/nucleon

Explanation:

mass of ¹³⁶Ba = 135.905 amu

¹³⁶Ba contain 56 proton and 80 neutron

mass of proton = 1.00728 amu

mass of neutron = 1.00867 amu

mass of ¹³⁶Ba = 56 x  1.00728 amu + 80 x 1.00867 amu

                      = 137.10128 amu

mass defect = 137.10128 - 135.905

                    = 1.19628 amu

mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg

                     = 1.9858 x 10⁻²⁷ Kg

speed of light = 3 x 10⁸ m/s

binding energy,

E = mass defect x c²

E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²

E = 17.87 x 10⁻¹¹ J/atom

now,

binding energy per nucleon =\dfrac{17.87\times 10^{-11}}{136}

                                              = 0.1312 x 10⁻¹¹ J/nucleon

                                              = 1.312 x 10⁻¹² J/nucleon

4 0
3 years ago
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