Answer:
- 7.088 m/s²
Explanation:
As we know that,
★ Acceleration = Change in velocity/Time
→ a = (v - u)/t
Here,
- Initial velocity (u) = 27.7 m/s
- Final velocity (v) = 10.9 m/s
→ a = (10.9 m/s - 27.7 m/s)/2.37 s
→ a = -16.8/2.37 m/s²
→ <u>a</u><u> </u><u>=</u><u> </u><u>-</u><u>7</u><u>.</u><u>0</u><u>8</u><u>8</u><u> </u><u>m/s²</u> [Answer]
Negative sign denotes that the velocity is decreasing.
Let's see: frequency of cellular phone waves (GSM phones) is (800-1900 MHz). If we look at the table of the electromagnetic spectrum, we can see that this range is contained within the frequencies of the microwaves, which include waves in the range 300 MHz-300 GHz.
So, summarizing, the correct answer is "microwaves".
Answer:
It traveled 4 centimeters.
Explanation:
In a speed versus time graph, the distance travelled is given by the area under the graph.
In this graph we have the following:
- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s
- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s
Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as

Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as

where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then

Resolving for F,

PART B) We need here to apply the shear strength equation, then



In such a way that the material is more resistant to tensile strength than shear force.