Answer:
1/3
Explanation:
We can solve the problem by using the lens equation:
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
Here we have a divering lens, so the focal length must be taken as negative (-f). Moreover, we know that the object is placed at a distance of twice the focal length, so
So we can find q from the equation:
Now we can find the magnification of the image, given by:
Answer:
The skidding distance would be doubled
Explanation:
When the truck applies the brakes and slows down, its motion is a uniformly accelerated motion, so its skidding distance can be found by using the suvat equation
where
v = 0 is the final velocity (zero since the truck comes to a stop)
u is the initial velocity
a is the acceleration
s is the skidding distance
The acceleration can also be written as
where F is the force applied by the brakes and m the mass of the truck. Substituting into the previous equation,
We see that the skidding distance is proportional to the mass: therefore, if the mass of the truck is doubled, the skidding distance will double as well.
Answer:
78g
Explanation:
Given parameters:
Mass of oxygen gas = 16g
Mass of potassium oxide = 94g
Unknown:
Mass of reacting potassium = ?
Solution:
To solve this problem, we need to obtain a balanced reaction equation. Then determine the number of moles of the reactant and use it to find that of the other one.
Balanced equation:
4K + O₂ → 2K₂O
Number of moles of reacting oxygen;
Number of moles =
molar mass of O₂ = 2 x 16 = 32g/mole
Number of moles = = 0.5mole
From the reaction equation;
4 mole of K reacted with 1 mole of O₂;
x mole of K will react with 0.5 mole of O₂
Therefore, 4 x 0.5 = 2 moles of potassium.
Mass of potassium = number of moles x molar mass
Molar mass of potassium = 39g
Mass of potassium = 2 x 39 = 78g