Answer:
x = 1.26 sin 3.16 t
Explanation:
Assume that the general equation of the displacement given as
x = A sinω t
A=Amplitude ,t=time ,ω=natural frequency
We know that speed V

V= A ω cosωt
Maximum velocity
V(max)= Aω
Given that F= 32 N
F = K Δ
K=Spring constant
Δ = 0.4 m
32 =0.4 K
K = 80 N/m
We know that ω²m = K
8 ω² = 80
ω = 3.16 s⁻¹
Given that V(max)= Aω = 4 m/s
3.16 A = 4
A= 1.26 m
Therefore the general equation of displacement
x = 1.26 sin 3.16 t
Answer:
Yes, I think
Explanation:
Melting is a process that causes a substance to change from a solid to a liquid. Melting occurs when the molecules of a solid speed up enough that the motion overcomes the attractions so that the molecules can move past each other as a liquid.
The correct answer is A.
The cell membrane consists of a phospholipid bilayer with embedded proteins. Sometimes molecules are just too big to easily flow across the plasma membrane or dissolve in the water so that they can be filtered through the cell membrane. In these cases , the cells must put out a little energy to help get molecules in and out of the cell.
The proteins embedded in the plasma membrane form channels through which other molecules can pass. Some proteins act as carriers, that is they are 'paid" in energy to let a molecule attach to itself and then transport that molecule inside the cell. This is called active transport.
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
brainly.com/question/2004529
#SPJ4