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Slav-nsk [51]
2 years ago
12

helicopter flies an average velocity of 70 m/s2 for 10.0 seconds. What ios ther magnitude of the helicopter's displacement? answ

er should have 3 significant figures
Physics
2 answers:
Inessa [10]2 years ago
7 0

The uniform movement allows finding the results for the helicopter displacement during the 10 s is: 700 m

Kinematics studies the motion of bodies, finding relationships between the position, velocity and acceleration of the body, in the special case that the acceleration is zero, it is called uniform motion and is described by the equation

          v = \frac{\Delta s}{t}

Where v is the average velocity, Ds is the displacement and t is the time

          Δs = v t

They indicate that the average speed is 70 m / s and the travel time is t=10.0s

Let's calculate

          Δs = 70 10.0

          Δs = 700 m / s

In conclusion using the uniform movement we can find the results for the helicopter displacement is: 700 m

Learn more here: brainly.com/question/14355103

mariarad [96]2 years ago
7 0

The magnitude of the helicopter's displacement is 700 meters.

<u />

<u />

<u>Given the following data:</u>

  • Velocity = 70 m/s
  • Time = 10 seconds.

To find the magnitude of the helicopter's displacement:

Velocity can be defined as the rate of change in displacement (distance) with time.

Also, velocity is a vector quantity and as a result of this, it has both magnitude and direction.

Mathematically, velocity is given by the formula;

Velocity = \frac{displacement}{time}

Making displacement the subject of formula, we have:

Displacement = velocity × time

Displacement = 70 × 10

<em>Displacement </em><em>=</em><em> 700 meters.</em>

Therefore, the magnitude of the helicopter's displacement is 700 meters.

Read more: brainly.com/question/17742679

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The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu
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6.66\cdot 10^{-12}T

Explanation:

The magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

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5. When an object reaches terminal velocity its acceleration is
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A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is o
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Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

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= E∫ds  ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

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= 0.05 V/0.04 m

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= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

7 0
2 years ago
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