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3 years ago

Which of the followijg is an example of kinetic energy

Physics

2 answers:

andrezito [222]3 years ago

4 0

Explanation:

The answer to ur question is option D) A swing moving back n forth bcos kinetic energy is energy that is due to motion....even tho option C) states that the rubberband is stretched, it is still considered a potential enrgy bcos it is in a position that wud change from potential to kinetic energy....sp the correct ans iz option D)

Ipatiy [6.2K]3 years ago

3 0

Answer:

Explanation:

D. A swing moving back and forth

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You toss a conductive open ring of diameter d = 1.75 cm up in the air. The ring is flipping around a horizontal axis at a rate o

Mama L [17]

Answer:

The maximum emf induced in the ring

= (2.882 × 10⁻⁷) V

Explanation:

According to the law of electromagnetic induction, the emf induced in the ring is given by

E = N BA w sin wt

The maximum emf induced is

E = N BA w

B = 30.5 μT = (30.5 × 10⁻⁶) T

A = (πD²/4)

D = 1.75 cm = 0.0175 m

A = (π×0.0175²/4) = 0.000240625 m²

Nw = 2π × 6.25 = 39.29 rad/s

E = 30.5 × 10⁻⁶ × 0.000240625 × 39.29

E = (2.882 × 10⁻⁷) V

Hope this Helps!!!

8 0

3 years ago

Use what you know about reflection and absorption of light to complete the sentence.

Blababa [14]

A red ladybug appears red in white light, red in red light, and black in blue light. Those would be the proper selections you'd need.

3 0

3 years ago

Read 2 more answers

Find the velocity in m/s of a swimmer who swims 110m toward the shore in 72s

ruslelena [56]

S=Vt
110=V(72)
110/72=V
V=1.527m/s

4 0

2 years ago

a 2 meters tall person is located 5 meters from a camera lens (camera lens are convex lenses). the lens has a focal length of 35

ki77a [65]

Sorry that you got your answer late but the answer is 0.035m

3 0

3 years ago

An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,

Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity

$v=\sqrt{2.14^2+0.34^2}=2.17m/s$

Direction,

$\theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0$

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0

3 years ago