Yellow and red dwarf stars are normal stars - they burn hydrogen in their cores and live on the main sequence of stellar lifetimes.
Answer:
50km /hour
Explanation:
The car moves 100km in two hours.
We can write a ratio like this:
100km : 2hr
Divide both sides by two and you get:
50km : 1hr
Meaning the moving car is moving 50km an hour.
Hope this helps :)
Answer:
1. 0 J
2. 7500 J
3. 7500 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Final velocity (v₂) of car = 5 m/s
Original kinetic energy (KE₁) =?
Final kinetic energy (KE₂) =?
Work used =?
1. Determination of the original kinetic energy.
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Original kinetic energy (KE₁) =?
KE₁ = ½mv₁²
KE₁ = ½ × 600 × 0²
KE₁ = 0 J
Thus, the original kinetic energy of the car is 0 J.
2. Determination of the final kinetic energy.
Mass (m) of car = 600 Kg
Final velocity (v₂) of car = 5 m/s
Final kinetic energy (KE₂) =?
KE₂ = ½mv₂²
KE₂ = ½ × 600 × 5²
KE₂ = 300 × 25
KE₂ = 7500 J
Thus, the final kinetic energy of the car is 7500 J
3. Determination of the work used.
Original kinetic energy (KE₁) = 0
Final kinetic energy (KE₂) = 7500 J
Work used =?
Work used = KE₂ – KE₁
Work used = 7500 – 0
Work used = 7500 J
Answer:
length = 7*10^(-8)km
width = 4.666*10^(-8) km
Explanation:
We know that:
1 μm = 1*10^(-6) m
and
1km = 1*10^3 m
or
1m = 1*10^(-3) km
if we replace the meter in the first equation, we get:
1 μm = 1*10^(-6)*1*10^(-3) km
1 μm = 1*10^(-6 - 3)km
1 μm = 1*10^(-9)km
Now with this relationship we can transform our measures:
Length: 70 μm is 70 times 1*10^(-9)km, or:
L = 70*1*10^(-9)km = 7*10^(-8)km
And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:
W = 46.66*1*10^(-9)km = 4.666*10^(-8) km
