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Korvikt [17]
3 years ago
6

Challenge: In a 60 km/h speed zone the police set up a digitector, placing 2 cables across the road, 5m apart The time taken for

cars to cross both cables was recorded. One car took 0.2 seconds to cross the cables. Was it speeding? If so by how much?​
Chemistry
1 answer:
creativ13 [48]3 years ago
7 0

Answer:

The answer to your question is given below.

Explanation:

The following data were obtained from the question:

Speed limit = 60 km/h

To know if the car was speeding, let us calculate the speed of the car.

This can be obtained as follow:

Distance (d) = 5 m

Convert 5 m to km

Recall:

1000 m = 1 km

5 m = 5/1000

5 m = 0.005 km

Time (t) = 0.2 sec.

Convert 0.2 sec to hour.

Recall:

3600 secs = 1 h

Therefore,

0.2 sec = 0.2/3600

0.2 sec = 5.56×10¯⁵ h

Thus,

Distance (d) = 0.005 km

Time (t) = 5.56×10¯⁵ h

Speed of car =?

Speed = Distance /time

Speed = 0.005/5.56×10¯⁵

Speed of the car = 90 km/h

Comparing the speed of the car (i.e 90 km/h) to the speed limit of the zone (i.e 60 km/h), we can see that the car is on a high speed since its speed (i.e 90 km/h) is higher than the speed limit of the zone (i.e 60 km/h).

Speed of the car = 90 km/h

Speed limit = 60 km/h

Difference = 90 – 60

Difference = 30 km/h

Therefore, the car is speeding by 30 km/h more compared to the speed limit.

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Determine the number of significant figures in the following measurement: <br><br> 16.08 Liters
valentina_108 [34]

Answer:

Four significant figures.

Explanation:

Hello,

In this case, considering that the measurement 16.08 L has four digits we must take into account that all are significant even the zero due to fact that is to the right of the first nonzero digit which is 1, therefore, the measurement has four significant figures.

Best regards.

7 0
3 years ago
Ammonia (NH3) is widely used as a fertilizer
ratelena [41]

Answer:

82.28g

Explanation:

Given parameters:

Number of moles of hydrogen gas = 7.26 moles

Unknown:

Amount of ammonia produced = ?

Solution:

We have to write the balanced equation first.

              N₂    +   3H₂    →    2NH₃

Now, we work from the known to the unknown;

   

                  3 moles of H₂ will produce 2 moles of NH₃

                7.26 mole of H₂ will produce \frac{7.26 x 2}{3}  = 4.84 moles of NH₃

Molar mass of NH₃ = 14 + 3(1) = 17g/mol

Mass of NH₃  = number of moles  x molar mass = 4.84 x 17 = 82.28g

3 0
3 years ago
What is the difference between constitutional isomers and confirmational isomers?
babymother [125]

Answer:

<h3>Constitutional Isomers are two molecules with the same composition but different constitution. Confirmational Isomers are two molecules with the same configuration but different confirmation.</h3>

Explanation:

4 0
3 years ago
Its atoms contain 3 protons and 4 neutron(s).<br> Express your answer as an isotope.
velikii [3]

Answer: A=7 for the isotope with 4 neutrons.

Explanation:A lithium atom contains 3 protons in its nucleus. Notice that because the lithium atom always has 3 protons, the atomic number for lithium is always Z = 3. The mass number, however, is A = 6 for the isotope with 3 neutrons, and A = 7 for the isotope with 4 neutrons.

3 0
3 years ago
A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 8.5 oz . aluminum has a density of 2.70 g/cm3. w
Serhud [2]
Hello!

We have the following data:

Area (A) = 50 square feet
Mass (m) = 8.5 ounces
Density (d) = 2.70 g/cm³
Volume (V) = ?
Thickness (T) =? (in mm)

To move on, we must transform the area of 50 ft² in cm², let's see:

1 ft² ------- 929,0304 cm²
50 ft² ----- A

A = 50*929,0304

\boxed{A = 46451,52\:cm^2}\Longleftarrow(Area)

In the same way, we will convert the mass of 8.5 oz in grams, see:

1 oz -------- 28,3495 g
8,5 oz ------- m

m = 8,5*28,3495

\boxed{m = 240,97075\:g}\Longleftarrow(mass)

Knowing that the density is 2.70 g/cm³ and the mass is 240.97075 g, we will find the volume, applying the data in the density formula we have:

d =  \dfrac{m}{V} \to V =  \dfrac{m}{d}

V =  \dfrac{240,97075\:\diagup\!\!\!\!\!g}{2,70\:\diagup\!\!\!\!\!g/cm^3}

V = 89,24842593... \to \boxed{V \approx 89,25\:cm^3}

The statement wants to find the thickness of the packaging, for this we have some important data, such as: V (volume) = 89,25 cm³ and Area (A) = 46451,52 cm² and T (thickness) =? (in mm)

In the calculations of Costs in Surface Treatment of a part within the flat geometry, we will use the following formula:

V (volume) = A (Area) * T (Thickness)

89,25\:cm^3 = 46451,52\:cm^2\:*\:T

46451,52\:cm^2*T = 89,25\:cm^3

T =  \dfrac{89,25\:cm^3}{46451,52\:cm^2}

T = 0,001921358009...\:cm

We will convert to millimeters, going through a decimal place on the right

T = 0,01921358009..\:mm

\boxed{\boxed{T \approx 0,0192\:mm\:or\:T\approx \:1,92*10^{-2}\:mm}}\end{array}}\Longleftarrow(thickness)\qquad\checkmark

Hope this helps! :))





8 0
3 years ago
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