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:
(a) coefficient of friction = 0.451
This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)
(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.
Explanation:
The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.
Thank you for reading and I hope this is helpful to you.
Answer:
The density of the fluid is 1100 kg/m³.
Explanation:
Given that,
Height = 5.00 cm
Pressure at top =594 Pa
Pressure at bottom = 1133 Pa
We need to calculate the change in pressure
Using formula of change in pressure
Where, 
= Pressure at bottom
= Pressure at top
put the value into the formula
Using formula of pressure for density
Where, 
= density
P = pressure
h = height
Put the value in to the formula
Hence, The density of the fluid is 1100 kg/m³.
Answer:C
Explanation:
Mass energy of hydrogen fusing into helium
Answer:
d = 3.54 x 10⁴ Km
Explanation:
Given,
The distance between the two objects, r = 2.5 x 10⁴ Km
The gravitational force between them, F = 580 N
The gravitational force between the two objects is given by the formula
F = GMm/r² newton
When the gravitational force becomes half, then the distance between them becomes
Let us multiply the above equation by 1/2 on both sides
( 1/2) F = (1/2) GMm/r²
= GMm/2r²
= GMm/(√2r)²
Therefore, the distance becomes √2d, when the gravitational force between them becomes half
d = √2r = √2 x 2.5 x 10⁴ Km
= 3.54 x 10⁴ Km
Hence, the two objects should be kept at a distance, d = 3.54 x 10⁴ Km so that the gravitational force becomes half.
