Water is boiled at 1 atm pressure in a 25-cm-internal- diameter stainless steel pan on an electric range. if it is observed that
1 answer:
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Refer to the diagram shown below.
g = 9.8 m/s², and air resistance is ignored.
For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.
For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.
Let a = the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²
Answer: 0.665 m/s² (nearest thousandth)
g = 9.8 m/s², and air resistance is ignored.
For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.
For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.
Let a = the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²
Answer: 0.665 m/s² (nearest thousandth)
Answer:
South = 1.5m
West =4.2m
Explanation:
Kindly see attached a rough draft of the situation
Step one
Given data
From the sketch the direction of the player is along the resultant of the triangle, corresponding to the Hypotenuse
Step two:
Hence for an opponent to tackle him towards the south, he must be at
sin θ= opp/hyp
sin 20=x/4.5
x=sin 20*4.5
x=0.342*4.5
x= 1.5m
Also, for an opponent to tackle him towards the south, he must be at
cos θ= adj/hyp
cos 20=y/4.5
y=cos 20*4.5
y=0.93*4.5
y= 4.2m
