Very high melting<span> points - Substances with giant covalent structures have very high</span>melting<span> points, because a lot of strong covalent bonds must be broken. Graphite, for example, has a </span>melting point<span> of more than 3,600ºC. Variable conductivity - Diamond does not conduct electricity.</span>
It's called condensation. That is what that wetness on the outside of the cup is.
Answer:
Microwave: Radiation
Water: Conduction, convection & radiation.
Explanation:
- When we heat a food using microwave then the water content of the food is only heated by the microwave.
- Microwaves make the molecules of water vibrate frequently with a frequency closer to the frequency of microwaves and this increases the kinetic energy of the molecules which produces heat in the water molecules this heat then propagates to the whole food by conduction and convection, but the heat enters the food only via radiation.
Now when the food item is kept into warm water then the molecules closer to the food heat the food by conduction through direct energy transfer by lattice vibrations and when they become cooler than the mass of water all around then due to density difference they settle down and their place is occupied by warmer molecules around in the fluid leading to convection.
Radiation of energy from a mass occurs continuously irrespective of the medium present there. So the heat of water also enters the food by radiation of energy from the water molecules.
Ok? I don’t know what you want me to do though
Answer:
Explanation:
Given that
Height of the tree is 3.7m
Therefore yo=3.7m
yox= 0m
The tiger lands 4.8m from the tree
Then, Range x=4.8m
Since she flies horizontally and she lands away from the bottom of the tree, her path will be trajectory as shown in the attachment
Let know the time of flight, using the equation of motion
voy is the initial velocity of the vertical motion of the is the tiger, which is zero at the beginning.
y = y0 + voy*t + ½*g*t²
Given that,
y=3.7m
yo=0m
voy=0m/s
g=9.81m/s²
y = y0 + voy*t + ½*g*t²
3.7=0+0•t+½×9.81×t²
3.7=0+0+4.905t²
3.7=4.905t²
t²=3.7/4.905
t²=0.7543
t=√0.7543
t=0.87sec
Time to reach the ground is
Now to know the initial velocity of the horizontal motion, using equation of motion
x=xo+Voxt
Vox is the horizontal initial velocity of the tiger.
x=xo+Voxt
4.8=0+Vox×0.87
4.8=0+0.87Vox.
4.8=0.87Vox.
Then, Vox=4.8/0.87
Vox=5.52m/s
Then, her initial velocity Vo
Vo=√voy²+vox²
Vo=√0²+5.52²
Vo=√5.52²
Vo=5.52m/s
The initial velocity is 5.52m/s
