Answer:
Vmax=11.53 m/s
Explanation:
from conservation of energy

Spring potential energy =potential energy due to elevation
0.5*k*x²= mg
=mgh
0.5*k*2.3²= 430*9.81*6
k=9568.92 N/m
For safety reason
k"=1.13 *k= 1.13*9568.92
k"=10812.88 N/m
agsin from conservation of energy

spring potential energy=change in kinetic energy
0.5*k"*x²=0.5*m*
10812.88 *2.3²=430*
=11.53 m/s
The frictional force is given by F = μmg
<span>where μ is the coeficient of friction. </span>
<span>Work done by frictional force = Fd = μmgd </span>
<span>Kinetic energy "lost" = 1/2 mv² </span>
<span>Fd = μmgd = 1/2 mv² </span>
<span>The m's cancel μgd = v² / 2 </span>
<span>d = v² / 2μg </span>
<span>d = 8² / 2(0.41)(9.8) </span>
<span>d = 32 / (0.41)(9.8) </span>
<span>d = 7.96 </span>
<span>Player slides 8 m . </span>
<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>
<span>d = v² / 2μg </span>
<span>= 4² / 2(0.46)(9.8) </span>
<span>= 8 / (0.46)(9.8) </span>
<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D
Answer: 1.51 km
Explanation:
<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.
Or, 
Where Q1 and Q2 are magnitude of two charges and r is distance between them:
<u>Given:</u>
Q1 = Charge near top of cloud = 48.8 C
Q2 = Charge near the bottom of cloud = -41.7 C
Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N
k = 8.99 x 109Nm^2/C^2
<u>So,</u>

Therefore, the separation between the two charges (r) = 1.51 km
Answer:
Approximately
(given that the magnitude of this charge is
.)
Explanation:
If a charge of magnitude
is placed in an electric field of magnitude
, the magnitude of the electrostatic force on that charge would be
.
The magnitude of this charge is
. Apply the unit conversion
:
.
An electric field of magnitude
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.