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Pani-rosa [81]
2 years ago
9

Victoria has a crate of vegetables that weighs 100 newtons. She exerts a force of 100 newtons to lift the crate with a pulley. W

hat’s the mechanical advantage in this situation?
A. less than 1
B. equal to 1
C. equal to 0
D. more than 1
Chemistry
2 answers:
Dennis_Churaev [7]2 years ago
7 0

Simply put, MA = Force Out / Force in. That's the way it is usually stated. The force out is normally what you need to move. The force in is what you need to supply to get the force out. Most machines will give you an MA of more than 1. Some (like your arm) will give you less than 1 and others (like this one) will give you exactly one.

This one is frictionless, otherwise it would slip into less than one if it had friction.

Answer B


shutvik [7]2 years ago
4 0

Answer:

Mechanical advantage in this situation is equal to 1.

Explanation:

The mechanical advantage of a simple machine gives a relationship between the input force applied to the output force. Mathematically, it is given by :

MA=\dfrac{F_o}{F_i}

Where

F_o\ and\ F_i are output force and input force

In this case, the weight of the crate of vegetables is 100 N

Force exerted by her to lift the crate with a pulley is 100 N

The mechanical advantage in this situation is given by :

MA=\dfrac{100}{100}=1

So, the mechanical advantage in this situation is equal to 1. Hence, the correct option is (b) "equal to 1".          

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The highest principal enegy level of period 2 elements is 2. Period 3 elements all have six 3p electrons. Period 4 elements have
Dmitrij [34]

Explanation:

The highest principal energy level of period 2 elements is 2. (True)

The highest principle energy level of elements of any period is equal to period number. So the given statement is true.

Period 3 elements all have six 3p electrons – (False)  

For example, sodium and sulfur are period 3 elements but do not contain six 3p electrons. Only Argon of period possesses six p-electrons.

Period 4 elements have an inner electron configuration of [Ar]- (True)

Atomic number of elements belonging to period 4 have atomic number more than 18 (atomic number of Ar). So they must have an inner electron configuration of [Ar]

The valence electrons of group 5A elements are in the 6s subshell – [FALSE]

Elements of group 5A are nitrogen, arsenic, antimony and bismuth. They are p-block elements. Therefore, valence electrons are present in p-subshell.

Group 8A elements have full outer principal s and p subshells –(True)  

Group 8A elements are also known as noble gas element. Octets of noble gases are complete and hence, have full outer principal s and p subshells.

The valence electrons of group 2A elements are in an s subshell – (True)

Elements of group 2A are barium, magnesium, calcium, etc. They belong to s-block elements. Therefore, their valence electrons will be in s-subshell.

The highest principal energy level of period 3 elements is 4 – (False)

The highest principle energy level of elements of any period is equal to period number.

For example, sodium is present in 3rd period, so its principle quantum number will be 3.

Period 5 elements have an inner electron configuration of [Xe] – (False)

Period 5 elements have an inner electron configuration of [Kr]. The first member of this period is rubidium. 37 and that of Kr is 36.  

6 0
2 years ago
The metallic radius of a lithium atom is 152 pm. What is the volume of a lithium atom in cubic meters?
Anuta_ua [19.1K]

Answer:

Volume of lithium atom is found to be 1.47 X 10⁻²⁹ m³

Explanation:

Let us consider the volume of atom as a sphere (but it is little complex than that). This volume is mathematically expressed as,

V=\frac{4}{3}\pi  R^{3}----------------------------------------------------------------------------------------(Eq. 1)

Here, R is the radius of lithium atom. The radius is given in picometers, so firstly let us convert it into meters

R=(152pm )(1X10^{-12}\frac{m}{pm})

R = 1.52 X 10^{-10}m

placing this value in Eq.1 the required result is achieved

V=\frac{4}{3}\pi  {1.52X10^{-10}}^{3}

V= 1.47 X 10⁻²⁹ m³

8 0
2 years ago
What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c
brilliants [131]

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and  (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

8 0
2 years ago
Read 2 more answers
If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct
MA_775_DIABLO [31]

Answer:

1. sp  = XY₂

2. sp²  = XY₂Z, XY₃

3. sp3³ = XY₄, XY₂Z₂, XY₃Z

4. sp³d  = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

5. sp³d² = XY₆, XY₄Z₂, XY₅Z

Explanation:

this is quite dicey, so it should be looked into carefully.

we would classify each of the abbreviation according to their  hybridization and it electron domain.

⇒ sp hybridization = XY₂

in this, we can see that the central atom X is bonded to two outer atoms Y.

this makes the no of hybrid orbitals and the no of sigma bonds both 2.

electron domain = 2.

⇒ sp² hybridization = XY₂Z, XY₃

Here we can see the central atom X bonded with three outer atoms Y in XY₂Z and in XY₃. For XY₂Z molecule, the no of sigma bonds is 2 and the no of hybrid orbitals is 3. While for XY₃ molecule, the no of sigma bonds is 3 while the no of hybrid orbital is 3.

electron domain = 3.

⇒ sp³ hybridization = XY₄, XY₂Z₂, XY₃Z

for XY₄ molecule, the central atom X is bonded with four outer atoms Y. It has 4 numbers of both the sigma and orbital atoms.

In XY₂Z₂, the central atom X is bonded to 2 outer atoms Y, and has 2 lone pairs Z. From this, the no of hybrid orbitals is 4 and the no of sigma bonds is 2, with 2 lone pairs causing the sp³ hybridization.

⇒ sp³d hybridization = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

for all the molecules listed above, the sum of both the lone pairs and the outer atoms both give a total of 5, hence have the sp³d structure, viz;

XY₂Z₃:

total electron domain = 2+3 = 5  

XY₃Z₂:

total electron domain = 2+3 = 5

XY₄Z:

total electron domain = 1+4 = 5

⇒ sp³d² hybridization = XY₆, XY₄Z₂, XY₅Z

Same thing goes for the above molecule, where the sum of both the outer atoms and the lone pairs gives a total of 6 as can be seen in the example below.

XY4Z2:

total electron domain = 2+4 = 6

cheers, i hope this helps.

5 0
2 years ago
what property of the noble gases most likely prevented the gases from being readily/easily discovered?
Vlad [161]

These gases very rarely react, with others and also noble gases are odourless and colourless.

Explanation:

  • Noble gases will not react with anything so that is the reason why they are known as an inert gas.
  • Noble gases are present in group 18 on the periodic table and following the rule of the octet which is they completed their orbital by s2p6 which is the highest energy level.
  • Most elements are discovering through their reactivity with the other elements, commonly with oxygen. In the case of a noble gas, it is difficult for a scientist to work with the gases which have very less or no chemical property in terms of their reactivity.

7 0
2 years ago
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