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
Always here to help. Bring it!!!
The focal point of a concave mirror is halfway along the radius, therefore the radius would be 2•16= 32 cm
Answer:
α =18.75 rad/s²
Explanation:
Given that
Acceleration a = 0.15 g
We know that g =10 m/s²
a= 0.15 x 10 = 1.5 m/s²
d= 16 cm
Radius r= 8 cm
Lets take angular acceleration =α rad/s²
As we know that
a= α r
Now by putting the values
1.5 = α x 0.08
α =18.75 rad/s²
Answer:
Explanation:
1) True. The stored energy (U) is proportional to the electric field strength (E). The electric field strength decreases when a dielectric is introduced hence inserting a dielectric decreases U.
2) False. From the formula 
, capacitance is inversely proportional to distance hence if the distance is doubled, capacitance decreases.
3) False. As the distance between the electric field and the object increases, its electric field decreases.
4) False. If a dielectric is inserted, the plates are further separated. Q stays the same.
5) True. The electric field strength decreases when a dielectric is introduced and capacitance is inversely proportional to electric field hence Inserting a dielectric increases C
6) True. If a dielectric is inserted, the plates are further separated. Q stays the same.
7) True. When the distance is doubled, U increases
