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4 years ago

Select the correct answer.

Physics

1 answer:

jek_recluse [69]4 years ago

4 0

Answer:

The answer is C.) Lacks all 3 types of antigens.

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a sample of iron has the dimensions of 2cm times 3cm times 2cm. if the mass of this rectangular-shaped object is 94g, what is th

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The answer is 7.83. The three is repeating.

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3 years ago

Which best describes the way a sound wave is sent through the radio?

Anna35 [415]

Answer:

sound wave-electrical wave-radio wave

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3 years ago

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A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the

Ann [662]

Answer:

Explanation:

Before it hits the ground:

The initial potential energy = the final potential energy + the kinetic energy

mgH = mgh + 1/2 mv²

gH = gh + 1/2 v²

v = √(2g (H - h))

v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))

v ≈ 2.0 m/s

When it hits the ground:

Initial potential energy = final kinetic energy

mgH = 1/2 mv²

v = √(2gH)

v = √(2 * 9.81 m/s² * 0.42 m)

v ≈ 2.9 m/s

Using a kinematic equation to check our answer:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)

v ≈ 2.9 m/s

3 0

3 years ago

Which of the following is a correct equation for total energy?

sweet [91]

Answer:

Total energy=Kinetic energy+Potential energy

Explanation:

the total energy will be obtained from the energy of an object in motion and the energy stored in the object

8 0

3 years ago

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what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo

Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately 3.331 × 10³⁶ J

Reasons:

The parameters required for the question are;

Mass of the Earth, M = 5.97 × 10²⁴ kg

Radius of the Earth, R = 6.38 × 10⁶ m

The rotational period of the Earth, T = 24.0 hrs.

$The \ moment \ of \ inertia \ of \ uniform \ sphere \ is \ I = \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}$

Which gives;

$\mathbf{I_{Earth}} = \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}$

$\mathrm{The \ rotational \ kinetic \ energy \ is} \ E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}$

$\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}$

Therefore;

$\omega = \dfrac{2 \cdot \pi}{24} = \dfrac{\pi}{24}$

Which gives;

$\mathbf{E_{rotational}} = \dfrac{1}{2} \times 9.7202107 \times 10^{37} \times \left( \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}$

The rotational kinetic energy of the Earth, $E_{rotational}$ = 3.331 × 10³⁶ Joules

Learn more here:

brainly.com/question/13623190

The moment of inertia from part A of the question (obtained online) is that of the Earth approximated to a perfect sphere.

Mass of the Earth, M = 5.97 × 10²⁴ kg

Radius of the Earth, R = 6.38 × 10⁶ m

The rotational period of the Earth, T = 24.0 hrs

3 0

3 years ago