1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Korvikt [17]
3 years ago
11

According to the law of reflection, what is the angle of incidence

Physics
2 answers:
eimsori [14]3 years ago
4 0

Answer:

It is the angle the incident ray makes with a line drawn perpendicular to the viewer.

Explanation:

gradpoint

Sergio [31]3 years ago
3 0

Answer:

It is the angle the incident ray makes with a line drawn perpendicular to the viewer.

Explanation:

You might be interested in
On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth's atmosphere over Chelyabinsk, Russia, and expl
fredd [130]

Answer:

156.67 m/s

0.45676 times the speed of sound

Explanation:

Distance from the ground = 23.5 km = 23500 m

Time taken by the blast waves to reach the ground = 2\ minutes\ 30\ seconds=2\times 60+30=150\ s

Spedd of the wave would be

Speed=\dfrac{Distance}{Time}\\\Rightarrow v_b=\dfrac{23500}{150}\\\Rightarrow v-b=156.67\ m/s

The velocity of the blast wave is 156.67 m/s

v = Velocity of sound = 343 m/s

\dfrac{v_b}{v}=\dfrac{156.67}{343}\\\Rightarrow v_b=v\dfrac{156.67}{343}\\\Rightarrow v_b=0.45676v

The blast wave is 0.45676 times the speed of sound

7 0
3 years ago
Which of the following is a result of gravitational forces in the Solar System?
postnew [5]
The correct answer is d
3 0
3 years ago
Percent Yield Lab Report
Vlada [557]

Based on the data obtained from the reaction, the following conclusions can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

<h3>What is the percent yield of the reaction?</h3>

Equation of the reaction is given below:

  • 2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid  = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced =  mass/molar mass

  • molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

<h3>Percent yield</h3>
  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.0170

  • moles of MgO produced =  mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

<h3>Percent yield</h3>
  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546

6 0
2 years ago
How much force is required to move a tractor 60m, when using 250J of work?
lianna [129]

Answer:

Work done,W= 250J

Displacement , s = 60

We know that, Work done = Force x displacement

i.e , W = Fxs

250 J = F x 60m

F = 250/60

=4.16 N

Hence , 4.16 N of Force is applied on the body.

5 0
2 years ago
Read 2 more answers
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
raketka [301]

  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
2 years ago
Read 2 more answers
Other questions:
  • Use the graph to determine the object’s average velocity.
    9·1 answer
  • Consider a system to be one train car moving toward another train car at rest. When the trains collide, the two cars stick toget
    8·2 answers
  • Three mountain climbers set out to climb a mountain from the same altitude and all arrive at the same location at the top. Mount
    9·1 answer
  • PLEASE HURRY WILL MARK BRAINLIEST IF CORRECT
    14·1 answer
  • a person starts at a position of 2 meters and finishes at a postion of 25 meters. the trip takes 4.5 seconds. what is the person
    14·1 answer
  • A 60.0-kg man jumps 1.70 m down onto a concrete walkway. His downward motion stops in 0.025 seconds. If he forgets to bend his k
    15·1 answer
  • HIIIIIIIIIIIIIIIIIII WHAT IS 50X550
    7·1 answer
  • Suppose you have two point charges of opposite sign. As you move them farther and farther apart, the potential energy of this sy
    7·2 answers
  • If humans cannot see ultraviolet waves how can ultraviolet light be used to gather evidence of a crime
    10·2 answers
  • Motion is a change in<br><br> 1.Time<br> 2.Speed<br> 3.Position<br> 4.Velocity
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!