Answer:
<h2>
<em>6,142mm²</em></h2>
Explanation:
Given the dimension of a paper measured by a ruler as 7.4 cm wide and 8.3 cm long, the area of the paper is expressed using the area for calculating the area of a rectangle as shown;
Area of the piece of paper = Length * Width
Given length = 7.4cm
Length = 74mm (Since 10mm = 1cm)
Width = 8.3cm
Width (in mm) = 83mm
We converted to mm since the ruler used to measure has a division of 1mm.
Substituting the given values into the formula, we will have:
Area of the piece of paper = 74mm * 83mm
Area of the piece of paper = 6,142mm²
<em>Hence, the area of the piece of paper is 6,142mm²</em>
The distance it falls is given by
x = (1/2)at^2
where a = acceleration due to gravity = 9.8 m/s^2
x = (1/2)(9.8)(18)^2
x = 1587.6 m
The answer is 1587.6 meters
Answer:
D
Explanation:
The friction force is the weight force times the coefficient of friction. So diving by the coefficient gives you the weight force which is equivalent to the normal force.
Answer:
Option 10. 169.118 J/KgºC
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1.61 KJ
Mass of metal bar = 476 g
Specific heat capacity (C) of metal bar =?
Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:
1 kJ = 1000 J
Therefore,
1.61 KJ = 1.61 KJ × 1000 J / 1 kJ
1.61 KJ = 1610 J
Next, we shall convert 476 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
476 g = 476 g × 1 Kg / 1000 g
476 g = 0.476 Kg
Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1610 J
Mass of metal bar = 0.476 Kg
Specific heat capacity (C) of metal bar =?
Q = MCΔT
1610 = 0.476 × C × 20
1610 = 9.52 × C
Divide both side by 9.52
C = 1610 / 9.52
C = 169.118 J/KgºC
Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC
10 joules of work is done by the object
