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Juli2301 [7.4K]
3 years ago
8

Look at jackie's planned workout. Based on her entries, what would she do to maintain an effective long term workout routine

Physics
1 answer:
ivann1987 [24]3 years ago
3 0
What is jackie's planned workout? .-.
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1. First blank is A. Conductors
Second blank is D. Insulators

2. C. Heat
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A ball of mass 8 kg falls from rest from a height of 100 m. Neglecting air resistance,calculate its kinetic energy after falling
astra-53 [7]
As an object falls from rest, its gravitational energy is converted to kinetic energy

G.P.E = K.E = mgh

K.E = (80 Kg)(9.8 m/s²)(30 m)

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4 years ago
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Which properties do metalloids share with metals?
Inga [223]
Metals are not brittle so it can’t be the first one or the third one, both metalloids and metals are shiny so it can’t be the second one. Therefore, it would be the last one because both metalloids and metals are shiny and both are solids at room temperature because it is not a high enough melting point.

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3 years ago
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Unpolarized light of intensity I=10 falls on two successive polarizer wheels with the angle between the polarizer wheels e-60°.
yan [13]

Answer:

I=\frac{10}{4}

Explanation:

A polarizer changes the orientation of the oscillations of a light wave.

I₀ = Intensity of unpolarized light = 10

θ = Angle given to the polarizer = 60°

Intensity of light

I = I₀cos²θ

⇒I = 10cos²60

\\\Rightarrow I=10\times \frac{1}{2}\times \frac{1}{2}\\\Rightarrow I=\frac{10}{4}

So, the after passing through the second polarizer is \mathbf{\frac{10}{4}}

5 0
3 years ago
A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other
serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, \omega_{rotation} = Angle turned/Time

∴ \omega_{rotation}  = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, \omega _{Star}, is given as follows;

\omega _{Orbit}  = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s.

3 0
3 years ago
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