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MakcuM [25]
2 years ago
7

Think about a typical school day. In the space provided below, describe how each of the different forms of energy we have learne

d about (chemical, electrical, heat, kinetic, potential, and solar) are used during a typical school day and how each form of energy is transformed into another.
Physics
2 answers:
krok68 [10]2 years ago
6 0

Answer:

Explanation:

Chemical energy- A bunsen burner burning a beaker filled with water.

Heat energy- The water in the beaker absorbing the heat from the burner.

Electrical energy- Running Fans and lights in a classroom by switches.

Solar energy- Solar energy harnessed by solar panels to run the fans and lights by converting it into electrical energy.

Potential energy- A ball being held by a student at a certain height possesses energy due to gravity.

Kinetic energy- The same ball being left by the boy from a certain height produces kinetic energy

Nezavi [6.7K]2 years ago
3 0

Answer:

During a typical school day all forms of eneergy is being utilised and also transfer of energy takes place from one form to another.

Explanation:

Chemical energy- A bunsen burner burning a beaker filled with water.

Heat energy- The water in the beaker absorbing the heat from the burner.

Electrical energy- Running Fans and lights in a classroom by switches.

Solar energy- Solar energy harnessed by solar panels to run the fans and lights by converting it into electrical energy.

Potential energy- A ball being held by a student at a certain height possesses energy due to gravity.

Kinetic energy- The same ball being left by the boy from a certain height produces kinetic energy

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The anwser is A or D
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2 years ago
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As a train accelerates uniformly, it passes successive 5-kilometer markers while traveling at velocities 10 m/s and 25 m/s. what
gregori [183]
<span>122.0 km/hr. First let’s make sure all of our units are in the base meter form: i.e. convert 5km to 5000m. (We will convert back to km later). The first thing to do is look at the equation relating velocity, acceleration, and distance: Vf^2 = Vi^2 + 2*a*d, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is distance. 25^2 = 10^2 + 2*a*5000 =?> 625 = 100 +10000a => a= 0.0525m/s^2. Now that we have acceleration, we can use the same equation again with different numbers.: Vf^2 = Vi^2 + 2*a*d = 25^2 + 2*0. 0525m*5000 = 625 + 525 =1150 => Vf^2 = 1150 => 33.9m/s. Convert to km/hour: 33.9m/s * 1km/1000m *60s/1min * 60min/ 1 hr = 122.0 km/hr.</span>
8 0
3 years ago
What is the final velocity, in meters per second, of a freight train that accelerates at a rate of 0.085 m/s2 for 7.5 min, start
Llana [10]

Answer:

v_f=41.65\frac{m}{s}

Explanation:

The final velocity is given by the following kinematic equation:

v_f=v_0+at

Here, v_0 is the initial velocity, a is the body's acceleration and t is the motion time. We have to convert the time to seconds:

7.5min*\frac{60s}{1min}=450s

Now, we calculate the final velocity:

v_f=3.4\frac{m}{s}+(0.085\frac{m}{s^2}(450s))\\v_f=41.65\frac{m}{s}

4 0
3 years ago
A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w
barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

Moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}

Moles\ of\ AgBr= 0.1225\ mol

The reaction taking place is:

MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4

From the reaction,

4 moles of AgBr is produced when 1 mole of MBr_4 undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of MBr_4 undergoes reaction

0.1225 mole of AgBr is produced when \frac {1}{4}\times 0.1225 mole of MBr_4 undergoes reaction

Moles of MBr_4 got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of MBr_4 = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

0.030625 = \frac{12.5843}{319.616 + x}

Solve for x,

we get, x = 91.2999 g/mol

<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

8 0
2 years ago
Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp
kotegsom [21]

Answer:

m2  = 83.3 g

Explanation:

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m_1v_{i1} + m_2v_{i2} = m_2v_{f2}

as both sphere has same speed so v_{i2} = v_{i1}

m_2 = \frac{m_1}[\frac{v_f2}{v_{f1}}+1}

from conservation of kinetic energy principle we have

\frac{1}{2}m_1v^{2}_{i1} + \frac{1}{2}m_2v^{2}_{i2} = \frac{1}{2}m_2v^{2}_f2

v_{f1} = \sqrt {\frac{(m_1+m_2) v^2_i1}{m_2}

v_{f1} =  v_{i2}\sqrt {\frac{(m_1+m_2)}{m_2}

\frac{v_{f1}}{v_{i2}} =\sqrt {\frac{(m_1+m_2)}{m_2}

substituting this value in above equation to get m2 value

m_2 = \frac{m_1}{\sqrt {\frac{(m_1+m_2)}{m_2}+1}}

solving for m2 we  get

m2 = \frac{m_1}{3}

m_1 = 250 g

      =\frac{250}{3}

  m2  = 83.3 g

7 0
2 years ago
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