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3 years ago

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The resolving power of electron microscopes is much better than the resolving power of light microscopes because the wavelength

of electron beams is ________ than the wavelength of visible light.

1 answer:

Misha Larkins [42]3 years ago

6 0

Answer:

because the wavelength of electron beams is much shorter than the wavelength of visible light

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What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t

Gekata [30.6K]

Start by facing East. Your first displacement is the vector

<em>d</em>₁ = (225 m) <em>i</em>

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

<em>d</em>₂ = (350 m) <em>j</em>

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )

<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>

The net displacement is

<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃

<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>

and its magnitude is

|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0

3 years ago

What is the length of a spring that has 450J of potential energy and a spring constant of 650N/m?

11111nata11111 [884]

Answer:

Δx = 1.2 m

Explanation:

The CHANGE of spring length) (Δx) can be found using PS = ½kΔx²

Δx = √(2PS/k) = √(2(450)/650) = 1.17669... ≈ 1.2 m

The actual length of the spring is unknown as it varies with material type, construction method, extension or compression, and other variables we have no clue about.

4 0

3 years ago

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 <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are; </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

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

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, $E_{rotational}$ = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

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

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

3 0

3 years ago

A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3

Kamila [148]

Answer: B) 0.00337 m3.

Explanation:

Given data:

Mass of the ball = 10kg

Weight of the ball in air = 98N

Weight of the ball in water = 65N

Solution:

To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.

= 98 - 65 / 9.8

= 33 / 9.8

= 3.37kg

The volume of the ball is 3.37kg

The density of water is 1kg per Liter.

So 3.37 kg of water would have a volume of 3.37 Liters.

Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).

7 0

3 years ago

A ruler of length 0.30m is pivoted at its centre. Equal and opposite forces of magnitude 2.0N are applied to the ends of the rul

kow [346]

Answer:

0.3858 Nm

Explanation:

The torque of the couple is the dot product of the force vector and the couple vector from 1 end of the ruler to the center. This equals to the product of their magnitude times the cosine() of the angle made by their direction:

$T = \vec{F} \cdot \vec{s} = Fscos(50^0) = 2 * 0.3 * 0.643 = 0.3858 Nm$

6 0

3 years ago