Plsss answer the questionsplsss guys im struggling
1 answer:
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Answer:
Given that
For A weight Wt= 22.7 N
m₁ = 22.7/10 = 2.27 kg
Force alone inclined plane
Wt₁ = m₁ g sin θ
Wt₁ =22.7 sin 17.2°
Wt₁ = 6.7 N
For B weight Wt₂= 34.5-N
m₂ = 3.45 kg
coefficient of friction ,μ= 0.219
θ = 17.2 degree
The friction force on the block A
Fr= μ m₁ g cos θ
Fr= 0.216 x 22.7 x cos 17.2°
Fr= 4.68 N
Lets take acceleration of system is a m/s²
Tension = T
From Newtons law
Wt₂ - Wt₁ - Fr = (m₁ +m₂) a
34.5 - 6.7 - 4.68 = (2.27 + 3.45 ) a
a= 4.05 m/s²
Block B
Wt₂ - T = m₂ a
T = 34.5 - 4.05 x 3.45
T= 20.52 N
Answer:
the final temperature of the tea is 7.39⁰C.
Explanation:
Given;
mass of the tea, m = 375 g
specific heat capacity of the tea, C = 4.184 JJ/g°C
initial temperature of the tea, t₁ = 95°C
the final temperature of the tea, t₂ = ?
Energy lost by the refrigerator, Q = 137,460 J
The energy lost by the refrigerator is given by the following formula;
-Q = mc(t₂ - t₁)
-137,460 =375 x 4.184(t₂ - 95°C)
-137,460 = 1569(t₂ - 95°C)
Therefore, the final temperature of the tea is 7.39⁰C.