Answer: 330 m
Explanation:
The speed of sound is defined as the distance traveled by the sound wave in a especific time :
Where:
is the speed of sound
is half the time the sound wave travels since the person speaks, the sound wave hits the hill and then returns to the person again as an echo
is the distance between the person and the hill
So, we have to find :
Finally:
Answer:
what are the options for me
Answer: The plasma membrane is called a selectively permeable membrane as it permits the movement of only certain molecules in and out of the cells. Not all molecules are free to diffuse. If plasma membrane ruptures or breaks down then molecules of some substances will freely move in and out of the cells. Semipermeable membrane describes a membrane that allows some particles to pass through (by size), whereas the selectively permeable membrane "chooses" what passes through (size is not a factor). Selective permeability is a property of cellular membranes that only allows certain molecules to enter or exit the cell. This is important for the cell to maintain its internal order irrespective of the changes to the environment. For example, water, ions, glucose and carbon dioxide may need to be imported or exported from the cell depending on its metabolic activity. Similarly, signaling molecules may need to enter the cell and proteins may need to be released into the extracellular matrix. The presence of a selectively permeable membrane allows the cell to exercise control over the quantum, timing and rate of movement of these molecules.
Explanation: hope this helps you with your work/assingment! Have a wonderful day/night!
Answer:
hello your question lacks some data and required diagram
G = 77 GPa, т all = 80 MPa
answer : required diameter = 252.65 * 10-^3 m
Explanation:
Given data :
force ( P ) = 660 -N force
displacement = 15 mm
G = 77 GPa
т all = 80 MPa
i) Determine the required diameter of shaft BC
considering the vertical displacement ( looking at handle DC from free body diagram )
D' = 0.3 sin∅ , where D = 0.015
hence ∅ = 2.8659°
calculate the torque acting at angle ∅ of CD on the shaft BC
Torque = 660 * 0.3 cos∅
= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N
hello attached is the remaining part of the solution