Answer:
420000N
Explanation:
Given parameters:
Mass of the train = 5.6 x 10⁵kg
Acceleration = 0.75m/s²
Unknown:
Resultant force = ?
Solution:
According to newton's second law, force is the product of mass and acceleration;
Force = mass x acceleration
Resultant force that acts on the train is given below;
Force = 5.6 x 10⁵kg x 0.75m/s² = 420000N
Answer: 42
Explanation:
I will answer this in English.
We know that the apple needs 5 seconds to reach the ground.
Each floor of the building has a height of 2.88m.
Now, when we drop something, the only force acting on the object is the gravitational one, so the acceleration of the apple is:
a = -g
for the velocity, we integrate the acceleration over time, and as the apple is dropped, we do not have any initial velocity, so we do not have a constant in the integration:
v = -g*t
for the position we integrate again, now we have an initial height H, so the position is:
p = (-g/2)*t^2 + H
now the apple hits the ground when p = 0, so we can solve this equation to find H.
i will use g = 9.8m/s^2
0 = (-4.9m/s^2)*(5s)^2 + H
H = 122.5 m
now knowing H, we can divide it by the height of a floor in the building and get the number of the floor.
N = 122.5m/2.88m = 42.5
this means that the apple was dropped in the floor 42 (the 0.5 means that the apple was not right where the floor 42 starts, it was dropped around the middle of the floor 42)
E) Protons, neutrons, and electrons
Answer:
i) 21 cm
ii) At infinity behind the lens.
iii) A virtual, upright, enlarged image behind the object
Explanation:
First identify,
object distance (u) = 42 cm (distance between object and lens, 50 cm - 8 cm)
image distance (v) = 42 cm (distance between image and lens, 92 cm - 50 cm)
The lens formula,
Then applying the new Cartesian sign convention to it,
Where f is (-), u is (+) and v is (-) in all 3 cases. (If not values with signs have to considered, this method that need will not arise)
Substituting values you get,
i) 
f = 21 cm
ii) u =21 cm, f = 21 cm v = ?
Substituting in same equation
v ⇒ ∞ and image will form behind the lens
iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.
