<span>Force = Work done / distance = 4Nm / 2m = 2N</span>
Answer:
2m/s^2
Explanation:
Clculate the acceleration:
V = u +at
20m/s = 0 + a*10s
a = 20m//10s
a = 2m/s²
From the data given , it is not possible to calculate the displacement , because no direction of motion is given
But it is possible to calculate the distance travelled
Distance = ut + ½ *a*t²
distance = 0 + ½ * 2m/s * 10²s
distance = 100m
Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
Answer:
The number of bright fringes per unit width on the screen is,
Explanation:
If d is the separation between slits, D is the distance between the slit and the screen and
is the wavelength of the light. Let x is the number of bright fringes per unit width on the screen is given by :

is the wavelength
n is the order
If n = 1,

So, the the number of bright fringes per unit width on the screen is
. Hence, the correct option is (B).