Answer:
The volume of water that was in the kettle is 1170 
Explanation:
Given:
Power, P = 2.0 kW = 2000 W, Mass of stainless steel, 
= 710 g = 0.71 kg at temperature of 
Part A:
If it takes time, t = 3.5 minutes to reach boiling point of water 
, then from conservation of energy,
Total energy supplied by the burner = Total heat gained by the water and the stainless steel to rise from to
i.e. Pt = 
(100 - 20 ) + 
(100 - 20 )
= 
= 1.17 kg
where = 4200 J/Kgk (specific heat capacity of water), = 450 J/Kgk (specific heat capacity of steel)
But volume of water in the the kettle, v = 
∴ v = 1170
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Answer:
The linear velocity of the racquet at the point of contact with the ball is 6 m/s.
Explanation:
Given;
angular velocity of the racquet, ω = 12 rad/s
distance of strike, r = 0.5 m
The linear velocity of the racquet at the point of contact is given by;
V = ωr
V = (12)(0.5)
V = 6 m/s
Therefore, linear velocity of the racquet at the point of contact with the ball is 6 m/s.
Answer:
1.) Magnitude = 5596 N
2.) Direction = 60 degrees
Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N
Let us resolve the two forces into X and Y component
Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N
Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )
= 2000 + 2828.43
= 4828.43 N
The resultant force R will be
R = sqrt ( X^2 + Y^2 )
Substitutes the forces at X component and Y component into the formula
R = sqrt ( 2828.43^2 + 4828.43^2 )
R = sqrt ( 31313752.53 )
R = 5595.87 N
The direction will be
Tan Ø = Y/X
Substitute Y and X into the formula
Tan Ø = 4828.43 / 2828.43
Tan Ø = 1.707106
Ø = tan^-1( 1.707106 )
Ø = 59.64 degree
Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.
Answer:
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Explanation:
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