Explanation:
Sucrose is a disaccharide which is composed of fructose and glucose. Sucrose molecule has oxygen atoms bonded to hydrogen atoms (O-H bonds - Polar groups) on all ends of its double 6-Carbon ring. The areas near the oxygen atoms are slightly negative, and the areas near the hydrogen atoms are slightly positive that is, the O-H bonds are polar. They bond with the neighbouring Oxygen and Hydrogen atoms because of their
dipole - dipole attractions and hence hydrogen bonds are formed.
However, the covalent bonds within the molecule aren't broken. But rather, the hydrogen bonds holding the sucrose molecules in the crystalline lattice.
<h2>
Answer:</h2>
The rate of deceleration is -0.14
<h2>
Explanation:</h2>
Using one of the equations of motion;
v = u + at
where;
v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)
u = initial velocity of the boat = 25m/s
a = acceleration of the boat
t = time taken for the boat to accelerate/decelerate from u to v = 3 minutes
<em>Convert the time t = 3 minutes to seconds;</em>
=> 3 minutes = 3 x 60 seconds = 180seconds.
<em>Substitute the values of v, u, t into the equation above. We have;</em>
v = u + at
=> 0 = 25 + a(180)
=> 0 = 25 + 180a
<em>Make a the subject of the formula;</em>
=> 180a = 0 - 25
=> 180a = -25
=> a = -25/180
=> a = -0.14
The negative value of a shows that the boat is decelerating.
Therefore, the rate of deceleration of the speed boat is 0.14
Explanation:
When the wire is connected to a battery, the compass needle moves and changes its position. This happens because the needle magnetizes the copper wire, thus, creating a force.
While the current in the wire produces a magnetic field and exerts a force on the needle. The insulation on the wire becomes energized and exerts a force on the needle. Hence, the compass needle moves and changes its position.
In general, that's not possible, unless the three numbers relate to
very specific quantities.
For example, if the three numbers are the object's height, temperature,
and cost, then they are of no help at finding the object's velocity.
