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Mashcka [7]
2 years ago
11

A force p is 46.67N. if it inclined at an angle of 50°, find its horizontal component​

Physics
1 answer:
Ivanshal [37]2 years ago
8 0

Explanation:

F_{x} = 46.67 N(cos 50°) = 30.0 N

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The mass of Mars is 6.42 × 1023 kg. Its moon Phobos is 9378 kilometers away from Mars and has a mass of 1.06 × 1016 kg. What is
dangina [55]

Answer:

5.16 × 1021 N

Hope this helps!

4 0
2 years ago
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Elements in group 2 are all called alkaline earth metals. What is most similar about the alkaline earth metals? how many protons
Rom4ik [11]

Elements in group 2 are called alkaline earth metals, the most similarity about the alkaline metals is which chemical properties they have.

<h2>Further Explanation </h2><h3>Periodic table  </h3>
  • Periodic table is a table that contains elements arranged in columns called groups and rows called periods.
  • Elements are arranged based on physical and chemical properties such that elements in the same group will have similar physical and chemical properties.  
<h3>Chemical families  </h3>
  • Based on the chemical properties elements belong to a family of elements sharing similar chemical or physical characteristics.
  • Examples of common chemical families include; alkali metals, alkaline-earth metals, halogens and noble gases among others.
<h3>Alkaline-earth metals  </h3>
  • These are elements that are found in group 2 of the periodic table. Alkaline-earth metals include, Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium.  

Properties  of Alkaline-earth metals

  • Alkaline earth metals have a valence of two since they form ions by losing two electrons from their outermost energy levels.
  • Unlike the Alkali metals, the earth metals have a smaller atom size and are not as reactive compared to alkali metals.
  • Alkaline-earth metals are highly metallic and are good conductors of electricity.  
  • They react with water and steam to form metal hydroxide and metal oxides respectively  
  • They react with air to form metal oxides  
  • Reactivity of alkaline metals depends on the ease of losing electrons, thus the reactivity increases down the group as the number of energy levels increases.
  • Additionally, alkaline-earth metals have low electronegativities and low electron affinities  

Keywords: Chemical families, alkaline-earth metals, reactivity  

<h3>Learn more about:  </h3>
  • Chemical families: brainly.com/question/1358941
  • Alkaline-earth metals: brainly.com/question/8498732
  • Properties of alkaline-earth metals: brainly.com/question/11116789
  • Reactivity of metals: brainly.com/question/7101478

Level: High school  

Subject: Chemistry  

Topic: Periodic table and chemical families  

Sub-topic: Alkaline-earth metals  

4 0
2 years ago
Read 2 more answers
Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe
Otrada [13]
For this task we need to use Kepler's third law:

T = 2*pi*\sqrt{ \frac{ r^{3}}{G*M} }

where T is orbital period in seconds, r is Venus's semi-major axis, G is gravitational constant and M is mass of the sun.

Distance from earth to sun is 1AU so if we know earths period and distance from the sun we can calculate Venus period.
Te-earths period
Tv-venus period

\frac{Te}{Tv} =  \sqrt{ \frac{ Re^{3} }{ Rv^{3} } }

From the text we know that Re/Rv = 1/0.723

Which means that:
Tv =  \sqrt{0.723^3} * Te

Tv = 0.614 Te
5 0
2 years ago
An electrical drill has 100 J of chemical energy. Of the 100 J, 45 J are transformed into kinetic energy and 20 J are transforme
Rainbow [258]
The answer to this question is the thermal energy of the drill is 35 J
4 0
2 years ago
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At t=0, an object of mass m is at rest at x=0 on a horizontal, frictionless surface.
vivado [14]

Answer:

a)v=-T\dfrac{F_0}{m}(e^{-\dfrac{t}{T}} -1)

b)v=T\dfrac{F_0}{m}

Explanation:

Given that

F_x=F_0e^{-\dfrac{t}{T}}

We know that force F given as

F=  m a

a=Acceleration

m=mass

F=m\dfrac{dv}{dt}

F_0e^{-\dfrac{t}{T}}=m\dfrac{dv}{dt}

\dfrac{dv}{dt}=\dfrac{F_0}{m}e^{-\dfrac{t}{T}}

dv=\dfrac{F_0}{m}e^{-\dfrac{t}{T}}dt

By applying boundary conditions

\int_{0}^{v} dv=\int_{0}^{t}\dfrac{F_0}{m}e^{-\dfrac{t}{T}}dt

v=-T\left [\dfrac{F_0}{m}e^{-\dfrac{t}{T}} \right ]_0^{t}\\v=-T\dfrac{F_0}{m}(e^{-\dfrac{t}{T}} -1)

The expression of velocity will be

v=-T\dfrac{F_0}{m}(e^{-\dfrac{t}{T}} -1)

When the time become so long

e^{-\dfrac{t}{T}}=e^{-\dfrac{\infty }{T}}=0

Therefore the velocity

v=-T\dfrac{F_0}{m}(0 -1)

v=T\dfrac{F_0}{m}

6 0
2 years ago
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