The average velocity of the car for the whole journey is 69.57 km/h.
The given parameters:
- <em>Length of the road, L = 320 km</em>
- <em>Distance covered = 240 km at 75 km/h</em>
- <em>time spent refueling, t₂ = 0.6 hr</em>
- <em>Final velocity, = 100 km/hr</em>
The time spent by the before refueling is calculated as follows;
![t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7Bv%7D%20%5C%5C%5C%5Ct_1%20%3D%20%5Cfrac%7B240%7D%7B75%7D%20%5C%5C%5C%5Ct_1%20%3D%203.2%20%5C%20hours)
The time spent by the car for the remaining journey;
![t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr](https://tex.z-dn.net/?f=t_3%20%3D%20%5Cfrac%7B320%20-%20240%7D%7B100%7D%20%5C%5C%5C%5Ct_3%20%3D%200.8%20%5C%20hr)
The total time of the journey is calculated as follows;
![t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours](https://tex.z-dn.net/?f=t%20%3D%20t_1%20%2B%20t_2%20%2B%20t_3%5C%5C%5C%5Ct%20%3D%203.2%20%5C%20hr%20%5C%20%2B%20%5C%200.6%20%5C%20hr%20%5C%20%2B%20%5C%200.8%20%5C%20hr%5C%5C%5C%5Ct%20%3D%204.6%20%5C%20hours)
The average velocity of the car for the whole journey is calculated as follows;
![v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Btotal%20%5C%20distance%20%7D%7Btotal%20%5C%20time%7D%20%5C%5C%5C%5Cv%20%3D%20%5Cfrac%7B320%7D%7B4.6%7D%20%5C%5C%5C%5Cv%20%3D%2069.57%20%5C%20km%2Fh)
Learn more about average velocity here: brainly.com/question/6504879
If the distance around the equator is reduced by half, then the radius is also reduced by half.
Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = <em>4 </em>.
The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes <em>4 times what it is now</em>.
It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror
Answer:
The correct answer is -
A (the entire green box): Chemical Equation
B (the blue box): Reactants
C (the arrow): Reacts to Form
D (the number): Coefficient
E (the purple box): Products
Explanation:
The chemical reaction of burning methane and oxygen is as follows;
Here, the green part A is the chemical equation that includes various parts that are reactants B, methane, and oxygen, C is an arrow that indicates the formation of products.
2 is here coefficient that indicates the moles of the oxygen which forms carbon dioxide and water in box E is products