Complete each sentence with the correct type of lens.

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour

Serhud [2]

For the answer to the question above, first find out the gradient.

<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>

<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>

<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>

<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>

<span>y is the velocity of the car, x is the time.
</span>I hope this helps.

4 0

3 years ago

The table shows the percentage of carbon dioxide in the Earth’s atmosphere in the years 1800 and 2013. Calculate the difference

Natalka [10]

The difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800 is 0.06 Kg

<h3>Data obtained from the question</h3>

Year 1800 percent = 0.028%
Year 2013 percent = 0.040%
Mass of air = 500 Kg
Difference =?

<h3>How to determine the mass of CO₂ in 500 Kg in year 1800</h3>

Year 1800 percent = 0.028%
Mass of air = 500 Kg
Mass of CO₂ =?

Mass = percent × mass of air

Mass of CO₂ = 0.028% × 500

Mass of CO₂ = 0.14 Kg

<h3>How to determine the mass of CO₂ in 500 Kg in year 2013</h3>

Year 1800 percent = 0.040%
Mass of air = 500 Kg
Mass of CO₂ =?

Mass = percent × mass of air

Mass of CO₂ = 0.040% × 500

Mass of CO₂ = 0.2 Kg

<h3>How to determine the difference</h3>

Mass of CO₂ in year 1800 = 0.14 Kg
Mass of CO₂ in year 2013 = 0.2 Kg
Difference =?

Difference = mass in 2013 - mass in 1800

Difference = 0.2 - 0.14

Difference = 0.06 Kg

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7 0

2 years ago

A lamp is connected to the power supply.

Vika [28.1K]

Answer:

There are 45 turns in the secondary coil.

Explanation:

Given that,

Input potential of the lamp, $V_{in}=5\ V$

The output potential of the lamp, $V_{out}=1.5\ V$

Number of turns in primary coil, $N_P=150$

We need to find the number of turns needed on the secondary coil. We know that the ratio for a transformer is as follows :

$\dfrac{V_{out}}{V_{in}}=\dfrac{N_s}{N_P}\\\\N_s=\dfrac{V_{out}N_P}{V_{in}}\\\\N_s=\dfrac{1.5\times 150}{5}\\\\N_s=45\$

So, there are 45 turns in the secondary coil.

5 0

3 years ago

Need help with this question. Thirty points.

Anon25 [30]

$1)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12 mv^2 \\\\\implies m = \dfrac{2E_k}{v^2} = \dfrac{2\times 4500}{30^2}=10~ kg\\ \\2)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12mv^2 \\\\\implies v^2 = \dfrac{2E_k}m\\\\\implies v = \sqrt{\dfrac{2E_k}{m}}=\sqrt{\dfrac{2 \times 320}{\dfrac{20}{9.8}}}= 17.709 ~ ms^{-1}\\\\3)\\\\\text{Kinetic energy,}\\\\E_k=\dfrac 12 mv^2=\dfrac 12 \times 50 \times 10^2=2500 ~J\\\\\\$

$4)\\\\\text{Potential energy,}\\\\E_p =mgh = 5\times 9.8\times 1.5 = 73.5 ~J$

5 0

2 years ago

The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 k is connected. What is

yKpoI14uk [10]

We use the voltage division problem between the load resistance, amplifier output resistance as

$V_{out} = V_{amlifire} \times \frac{R_{load} }{R_{load} + R_{out} }$ .

Here, $V_{out}$ is the output voltage, $V_{amlifire}$ is the amplifier voltage, $R_{load}$ is the load resistance and $R_{out}$ is the amplifier output resistance.

Therefore,

$1-\frac{20}{100} = \frac{1 \ k\Omega }{1 \ k\Omega +R_{out} } \\\\ R_{out} = \frac{1 \ k\Omega }{0.8} -1 \ k\Omega =1250 \Omega -1000 \Omega =250 \Omega$ .

Thus, the amplifier output resistance is $250 \ \Omega$ .

4 0

3 years ago