Answer:
Approximately 
, assuming that 
.
Explanation:
Let 
denote the time required for the package to reach the ground. Let 
and 
denote the initial and final height of this package.
.
For this package:
- Initial height:
. - Final height:
(the package would be on the ground.)
Solve for , the time required for the package to reach the ground after being released.
.
.
Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at 
.) From calculations above, the package would travel forward at that speed for about 
. That corresponds to approximately:
.
Hence, the package would land approximately in front of where the plane released the package.
CORRECT ANSWER:
a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.
STEP-BY-STEP EXPLANATION:
The complete question from book is
According to Figure 9.6, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?
a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.
b- Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus.
c- Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule.
d- Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors.
e- None of the other answer options is correct.
The answer is d hope this helps
Answer:
it is essential that the charge on the plates are of the same magnitude, but in the opposite direction
Explanation:
The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.
To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction
This is so that the fields created by each plate can be added inside and subtracted from the outside of the plates
6.
A. 1575 - 1265 = 310J
B. KE 1/2 MV^2
V=√2·KE/M = √2(310)/12 V = 7.2mls
C. PE = 1265 = mgh
h= 1265/mg = 1265/(12)(92) h= 10.8m
7.
A. KE = 1/2 mv^2 0.5(5)(12)^2 KE = 360J
B. PE = mgh = (5)(9.8)(2.6) PE = 127.4J
C. ME = KE + PE = 360 + 127.4 ME = 487.4J
