Answer:
117.6°
Explanation:
The vertical component of a force directed at some angle α from the vertical is ...
F·cos(α)
We want the vertical components of the wolf's force (Fw) and Red's force (Fr) to total zero. So for some angle from vertical α, Red's force will satisfy ...
Fw·cos(25°) + Fr·cos(α) = 0
cos(α) = -Fw/Fr·cos(25°) ≈ -(6.4 N)/(12.5 N)·0.906308 ≈ -0.464030
α ≈ arccos(-0.464030) ≈ 117.6°
Red was pulling at an angle of about 117.6° from the vertical.
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<em>Additional comment</em>
That's about 27.6° below the horizontal.
- Height (h) = 10 m
- Density (ρ) = 1000 Kg/m^3
- Acceleration due to gravity (g) = 10 m/s^2
- We know, pressure in a fluid = hρg
- Therefore, the pressure exerted by a column of fresh water
- = hρg
- = (10 × 1000 × 10) Pa
- = 100000 Pa
<u>Answer</u><u>:</u>
<u>1000</u><u>0</u><u>0</u><u> </u><u>Pa</u>
Hope you could understand.
If you have any query, feel free to ask.
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
A wave is characterized by the cyclic occurrences of crests and troughs. Wavelengthis defined as the distance between two consecutive troughs or two crests and the Frequency is defined as the number of cycles that pass through a point per second
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
