Answer:
a) The perimeter of the rectangle is 29.4 centimeters.
b) The uncertainty in its perimeter is 0.8 centimeters.
Explanation:
a) From Geometry we remember that the perimeter of the rectangle (
), measured in centimeters, is represented by the following formula:
(1)
Where:
- Width, measured in centimeters.
- Length, measured in centimeters.
If we know that
and
, then the perimeter of the rectangle is:


The perimeter of the rectangle is 29.4 centimeters.
b) The uncertainty of the perimeter (
), measured in centimeters, is estimated by differences. That is:
(2)
Where:
- Uncertainty in width, measured in centimeters.
- Uncertainty in length, measured in centimeters.
If we know that
and
, then the uncertainty in perimeter is:


The uncertainty in its perimeter is 0.8 centimeters.
That's a formula that shows the relationship between three quantities ...
weight, mass, and acceleration. If you know any two of them, then you
can use this formula to find the one you don't know.
Examples:
==> I have a rock with 2 kilograms of mass.
The gravitational acceleration on Earth is 9.8 m/s² .
How much does my rock weigh on Earth ?
Weight = (mass) x (grav acceleration)
= (2 kg) x (9.8 m/s²)
= 19.6 newtons
(about 4.41 pounds)
==> My brother weighs 770 newtons (about 173 pounds) on Earth.
What is his mass ?
Weight = (mass) x (grav acceleration)
770 newtons = (mass) x (9.8 m/s²)
Divide each side
by 9.8 m/s²: 770 newtons / 9.8 m/s² = mass
78.57 kilograms = mass
==> When I went to the Moon, I took along my 2-kilogram rock.
I weighed my rock on the Moon.
It weighs 3.25 newtons (about 0.73 pound) there.
What is the gravitational acceleration on the Moon ?
Weight = (mass) x (grav acceleration)
3.25 newtons = (2 kg) x (acceleration)
Divide each side
by 2 kilograms: (3.25 newtons)/(2 kg) = acceleration
1.63 m/s² = grav acceleration on the Moon
Determinatio/Goal setting
Oceanic, Crust, Geolific, techtonic
Answer:
a. Yes, because the acceleration of the crate is 2.0 m/s².
Explanation:
Given
--- f
--- m
--- t
--- v
Required
Does the system support 
Yes, it does and this is shown below
The crate is initially at rest; so:

Using the first equation of motion

Substitute values for v, u and t


Make a the subject


Using 
Substitute values for F and m

Divide both sides by 3



In both cases:

<em>Hence, option (a) is correct.</em>