<span>Lets call F the friction force which will act horizontally backwards.
As you are travelling at a constant velosity horizontally there is no overall resultant force in this direction.
ie. the force you pull with will be equal to the friction force resisting you. (you will initially have to have pulled with a greater force than the friction to get the suitcase moving)
the value of your force pulling is 60 cos26.9 (horizontally) - you should have learnt about resolving forces.
this must be equal to F
so
F=60cos26.9
F=53.5N
hope this helps you
please mark this as brainliest answer</span>
Answer:
- Water gained: 10
- Iron lost: -10
Explanation:
Given: Hot iron bar is placed 100ml 22C water, the water temperature rises to 32C
To find: How much heat the water gain, how much heat did the iron bar lost
Formula:Q = change T x C x M
Solve:
<u>How much heat water gained</u>
Initial heat = 22, then rose to 32. To find how much heat the water gained, simply subtract the current heat by the initial heat.
32 - 22 = 10
The water gained 10 amounts of heat.
<u>How much heat Iron lost</u>
Current heat = 32, then dropped to 22. To find how much heat the Iron lost, simply subtract the initial heat by the current heat.
22 - 32 = -10
The Iron lost -10 amounts of water.
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
