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3 years ago

What compound name belongs in box (1) in Figure 3?

Physics

2 answers:

faltersainse [42]3 years ago

7 0

The compound name for box 1 should be sodium chloride.
The type of bond for box 3 should be covalent.

kodGreya [7K]3 years ago

5 0

The compound name which belongs to box[1] is known as Sodium chloride.

It is the common salt.

Before going to answer the second question, first we have to understand ionic and covalent bond.

The ionic bond is formed between electropositive metal and electronegative non metal. In this type of bond,the electropositive metal will lose electrons and forms cation where electronegative element will accpert electrons to form anion.The cations and anions are attracted to each other to form a compound.

The covalent bond is a type of bond in which sharing of electrons will occur to satisfy octet role which will result in the stability of the compound.

Hence the correct answer to the second question will be covalent.

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Write 3.5 seconds as milliseconds

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3.5 seconds is 3500 milliseconds.

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If you wanted to find a sample of fermium, which has an atomic number of 100, where would you look?

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Answer:

Explanation:

Fermium is a synthetic element with the symbol Fm and atomic number 100. It is an actinide and the heaviest element that can be formed by neutron bombardment of lighter elements, and hence the last element that can be prepared in macroscopic quantities, although pure fermium metal has not yet been prepared.[3] A total of 19 isotopes are known, with 257Fm being the longest-lived with a half-life of 100.5 days.

It was discovered in the debris of the first hydrogen bomb explosion in 1952, and named after Enrico Fermi, one of the pioneers of nuclear physics. Its chemistry is typical for the late actinides, with a preponderance of the +3 oxidation state but also an accessible +2 oxidation state. Owing to the small amounts of produced fermium and all of its isotopes having relatively short half-lives, there are currently no uses for it outside basic scientific research.

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2 years ago

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Choose the word that BEST completes the following sentence. Saturn’s rings have been assigned a letter, based on their location.

puteri [66]

Answer:

The correct answer would be Saturn's Cassini Division.

Explanation:

Read about it here.

https://caps.gsfc.nasa.gov/simpson/kingswood/rings/

Hope this helps! :)

5 0

3 years ago

Convert 3402kgm/s to 20000Newtons

oee [108]

The 3,402 has units of kg-m/s. That's momentum. The 20,000 has units of Newtons. That's force. Momentum and force are different physical things, and you can't convert them from one to the other.

The best I can do for you is something like this:

Let's say you have a moving object with 3,402 kg-m/s of momentum, and you want to STOP it completely. You want to stand in front of it and push back on it, hard enough and for long enough to CHANGE its momentum from 3,402 kg-m/s to zero.

Also ... there's a limit to how hard you can push. The most force you can exert is 20,000 Newtons.

The amount you'll change its momentum is called the impulse you give it. The quantity of impulse is (force) x (length of time you push on it).

So you need to keep pushing it back for (T seconds) long enough so that

(20,000 Newtons of force) x (T seconds) = 3,402 kg-m/s of momentum .

Divide each side of that equation by (20,000 Newtons). Then it says:

(T seconds) = (3,402 kg-m/s) / (20,000 Newtons)

T = 0.1701 second

And that's how you provide just enough impulse to stop the flying object ... push on it with 20,000 Newtons of force for exactly 0.1701 second, and it loses all its momentum and falls out of the air onto the ground at your feet.

This story is the closest I can come to anything that looks like "convert"ing momentum into force.

3 0

3 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

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3 years ago