Answer:
the answer is A because
from tate 4 dozen is 48 and from joe the sixth multiple of eight is 48
A. The acceleration during the slide is 6.86 m/s²
B. The time taken to slide until he stops is 1.2 s
<h3>How to determine the force of friction</h3>
- Mass (m) = 81.5 Kg
- Coefficient of friction (μ) = 0.7
- Acceleration due to gravity (g) = 9.8 m/s²
- Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 N
- Frictional force (F) =?
F = μN
F = 0.7 × 798.7
F = 559.09 N
<h3>A. How to determine the acceleration</h3>
- Mass (m) = 81.5 Kg
- Frictional force (F) = 559.09 N
- Acceleration (a) =?
a = F / m
a = 559.09 / 81.5
a = 6.86 m/s²
<h3>B. How to determine the time </h3>
- Initial velocity (u) = 8.23 m/s
- Final velocity (v) = 0 m/s
- Decceleration (a) = -6.86 m/s²
- Time (t) =?
a = (v – u) / t
t = (v – u) / a
t = (0 – 8.23) / -6.86
t = 1.2 s
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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).
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Is that a question? If it is not what its the question?
Answer
Correct answers are 1.an increase or decrease in pressure 2.an increase or decrease in energy
Explanation
All existing matter can undergo phase change it means they may transform from one state to another. phase change of a matter may occur due to change in energy and change in pressures.
for example there is an ice cube which is a solid and if we want to change it phase into liquid water for that purpose we have to supply some energy to ice cube it means we have to give some heat to ice cube. After supplying heat energy to the ice it will turn into water it means there is phase change from solid to liquid due to supplying the heat. Similarly we can change the liquid water into solid ice cube by taking heat energy(reducing temperature) from the liquid water .
In some cases the matter doesn't want to undergo phase transformation. For example, oxygen will solidify at -361.8 degrees Fahrenheit at standard pressure.But , it can change to solid state at warmer temperatures when the pressure is increased.