You have got two experimental results E_1 = 9.8 m/s^2 and E_1 = 9.97 m^2the perc

Question

You have got two experimental results E_1 = 9.8 m/s^2 and E_1 = 9.97 m^2the percentage uncertainty in your result. How long la a 10 m deck In millimeters ?. In centimeter?. In kilometers. Express the answer In scientific notation. A container holda 2000 kg of gasoline. What Is the mass of the gas in grams What la the density of a cylindrical rod whose radius is 5cm, length is 20 cm and mass is 20 gm. You have measured the length of the rod In Inches and also In cm. The length of the rod la measured In Inches (12 Inches) and In centimeters (32 cm). Calculate the percentage error considering the actual K Is 2.54 In this measurement In a Force table experiment, you have put the following forces: 200 gms, at) degree C 300 gm at 60 degree and 400 gm at 135 degree. If you are asked to find the resultant magnitude of the forces theoretically, What will be the resultant On September 11- 2001, a plane traveling north at 2001 miles/hr turns, east travels at 400 miles/and hit the world trade center. With what speed direction the plane hit the world trade center. When a certain force Is applied to a I kg standard mass and its acceleration (O m/s^2. When the same force is applied to another object its acceleration is 5 m/s^2. The mass of the object is: A torque Is applied to a rigid object always tend to produce: a linear acceleration, b) rotational equilibrium, c) precision, d) rotational accelera none of these.

Explanation / Answer

Part 1)

Uncertainty can be given in several ways

absolute error

Em = (E2 +E1) /2

Em = (9.8 +9.97) /2 = 9.885 m/s2

E = E1-Em = 9.885 – 9.8 = 0.09 m/s2

E = E2 – Em = 9.97 – 9.885 =0.09 m/s2

the absolute value is taken

Part 2

Problem reduction units

L =10 m (103 mm/1m) = 1 104 mm

L = 10 m (102 cm/ 1 m) = 1 103 cm

L = 10 m (1 Km/103 m) = 1 10-2 Km

Part 3

Reduce unit

2000 Kg ( 103gr/1 Kg) = 2 106gr

Part 4

V= A L

A= R2

V = R2 L

V = 52 20 =1570.80 cm³

= m/V

= 20/ 1570.80

= 1.27 10-2 gr/cm3

Part 5

Reduce inches to centimeters

12 inches (2.54 cm/1 inch) = 30.48 cm

Em = (30.48 + 32)/2

Em = 31.24 cm

e% = (Em- E1)/Em 100

e% = (31.24 -30.48)100 /31.24

e% = 2.4%

Part 6

F T ( c )

200 0

300 60

400 135

We can see from the table that increase is not linear, so we must find expression describing the data, a quadratic curve should describe the three pairs of values. Given the equation of this curve we can find the values of force knowing the system temperature.

Part 7)

We use the triangle Pitagoras

V = sqrt ( Vn2 + Ve2)

V = sqrt ( 2002 + 4002)

V= 447.21 miles/h

Part 8

We use Newton's second law

F = m a1

F = 1 10

F = 10 N

F= m a2

m = F/a2

m = 10/5

m = 2 Kg

Part 9)

= r x F = I

whereby the torque produces an angular acceleration (rotational)

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