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3 years ago

I AM TIMED !! 30 POINTS Which is the smallest possible particle of an element?

Physics

2 answers:

zimovet [89]3 years ago

4 0

Answer:

atoms

Explanation:

they are the smallest

melamori03 [73]3 years ago

3 0

The correct answer is Atoms!

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A wave has a speed of 30 m/s, a frequency of 6 Hz, and a wavelength of 5 m. If the wavelength remains constant, and the frequenc

mixer [17]

Answer: 60m/s

Explanation:

The wavespeed is the distance covered by the wave in one second. It is measured in metre per second, and represented by the symbol V

Wavespeed (V) = Frequency F x wavelength λ

i.e V = F λ

In the first case:

Wavespeed = 30 m/s

Frequency of sound = 6Hz

Wavelength = 5m

In the second case:

Wavespeed = ?

Frequency of sound = (2x 6Hz = 12Hz)

Wavelength = 5m (remains constant)

Apply V = F λ

Wavespeed = 12 Hz x 5m

Wavespeed = 60m/s

Therefore, when frequency is doubled, the speed is also doubled. Thus, the new speed of the wave is 60m/s

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4 years ago

9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku

Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately 33.23 m/s in a direction from the North of

approximately 9.18°.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

v₁ = -( $\frac{\sqrt{2} }{2}$ × 3.5)·i + ( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

v₂ = ( $\frac{\sqrt{2} }{2}$ × 4.0)·i - ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-( $\frac{\sqrt{2} }{2}$ × 3.5) - ( $\frac{\sqrt{2} }{2}$ × 4.0))·i + (( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5) + ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15))·j

v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

Red Skull's velocity relative to Captain America, v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

Red Skull appears to be moving West at 5.3 m/s and North at 32.8 m/s

The direction is $arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}$

Therefore;

Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

$|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23$ ·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ 33.23 m/s

Learn more here:

brainly.com/question/24430414

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3 years ago