The wavelengths for visible light rays correspond to which of these optionsA. About the size of a pen
B. About the size of a virus or large molecule
C. About the size of an amoeba
D. Smaller than an atom
E. About the size of a football field

Answers

Answer 1

Answer: So we want to know what does the wavelength of vidible light corresponds to. So the wavelength of visible light is around 400 - 700 nm or 400*10^-9 m to 700*10^-9 m. Virus size is around 20 - 400 nm. Large moleculer are around 100 nm to 200 nm. Atoms are around 10^-10 m and all the other options are too large. So the correct answer is B.

Answer 2

Answer:

Answer:

here's your answer

Explanation:

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An 880 kg cannon at rest fires a 12.4 kg cannonball forward at 540 m/s. What is the recoil velocity of the cannon? (Unit m/s)

Answers

Answer:

The recoil velocity vector of the cannon is $(7.609,0,0)(m)/(s)$

Explanation:

We can solve this problem by applying the Momentum Conservation Principle.

The principle of conservation of momentum states that when you have an isolated system with no external forces, we can use the following equation to calculate the final velocity of one object.

$m1.v1=m2.v2$ (I)

Where '' $m1$ '' and '' $v1$ '' are the mass and velocity of the first object.

And where '' $m2$ '' and '' $v2$ '' are the mass and velocity of the second object.

The momentum is a vectorial magnitude.

If we use the equation (I) with the data given :

$(880kg).v1=(12.4kg).(540(m)/(s))$

$v1=7.609(m)/(s)$

If we considered as negative the sense of the velocity vector from the cannonball, the cannon's velocity vector will have the same direction but opposite sense that the cannonball's velocity vector (It will be positive).

We can give it a vectorial character like this :

$v1=(7.609,0,0)(m)/(s)$

The velocity vector will be entirely in the x-axis.

Answer:

7.61 m/s backwards

Explanation:

Initial momentum = final momentum

0 = (880 kg) v + (12.4 kg) (540 m/s)

v = -7.61 m/s

The cannon's recoil is 7.61 m/s backwards.

As you increase the number of supporting strings, what happens to the force needed to lift the bottom block

Answers

the weight is spread across multiple different strings, allowing for each string to require dealing with a large amount of stress, because the stress is evenly distributed throughout each string. hope i helped :3

Which type of element seeks to gain electrons? Question 23 options: A. Non-Metals B. Metalloids C. Isotopes D. Metals E. Noble Gases

Answers

isotopes, each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element.

A stone is dropped from a bridge and hits the pavement below in two seconds. What is the velocity of the stone when it hits the pavement?A. 0.20 m/s
B. 19.6 m/s
C. 4.9 m/s
D. 9.8 m/s

Answers

We have: a = v/t
Here, t = 2 s [ Given ]
a = 9.8 m/s² [constant value for earth system ]

Substitute their values into the expression:
9.8 = v/2
v = 9.8 × 2
v = 19.6 m/s

In short, Your Answer would be Option B

Hope this helps!

i just did the quiz and so the only 2 options that made logical sense would be B (19.6 m/s) or D (9.8 m/s) the reason someone could get confused between these two is because 19.6 divided by 2 is 9.8 m/s so for each second it will go down 9.8 m/s but since it wants to know the velocity it would be 19.6 m/s

hope this helps

An object has a mass of 48g and a volume of 6cm cubed (like a little 3 above the m) Calculate the density of this object

Answers

   Density = (mass) / (volume)

    = (48 g) / (6 cm³)

      = (48 / 6) (g / cm³)

=    8 g/cm³

i hope this helps The density (d) of a material is its mass (m) divided by its volume (V). ... Calculatethe density of an object that has a mass of 25.0g and the volume is 10.5ml. ... 3.Calculate the mass for the following object. If you know that the object

A 1.7-kg block of wood rests on a rough surface. A 0.011-kg bullet strikes the block with a speed of 670 m/s and embeds itself. The bullet-block system slides forward 2.4 m before coming to rest.