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Hunter-Best [27]
2 years ago
10

The flow of electrons through a wire or any conductor is called____.

Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
5 0
Electricity. Ruler. 69. N.
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A trapeze artist, with swing, weighs 800 N; he is momentarily heldto one side by his partner using a horizontal force so that th
Tamiku [17]

Answer:

461.88 N

Explanation:

F_{g} = Weight of the swing = 800 N

T = Tension force in the rope

F = Horizontal force being applied by the partner

Using equilibrium of force in vertical direction using the force diagram, we get

T Cos30 = F_{g}\\T Cos30 = 800\\T = \frac{800}{Cos30} \\\\T = 923.76 N

Using equilibrium of force in horizontal direction using the force diagram, we get

F = T Sin30\\F = (923.76) (0.5)\\F = 461.88 N

6 0
2 years ago
(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down
slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

h=ut+\dfrac{1}{2}at^2

Here, u = 0 and a = g

h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0
3 years ago
Which of the following is a health problem associated with obesity in children?
mihalych1998 [28]
Risk of not being able to reduce their weight
5 0
3 years ago
Why does an energy transfer not always result in phase change?
mash [69]

Answer:

These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. "If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance

4 0
3 years ago
Read 2 more answers
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

5 0
3 years ago
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