Answer:
a) 
b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).
c) The material is peat, possibly.
d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.
Explanation:
Given:
- mass of aluminium,

- mass of water,

- initial temperature of the system,

- mass of copper block,

- temperature of copper block,

- mass of the other block,

- temperature of the other block,

- final equilibrium temperature,

We have,
specific heat of aluminium, 
specific heat of copper, 
specific heat of water, 
Using the heat energy conservation equation.
The heat absorbed by the system of the calorie-meter to reach the final temperature.



The heat released by the blocks when dipped into water:

where
specific heat of the unknown material
For the conservation of energy : 
so,


b)
This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).
c)
The material is peat, possibly.
d)
The material cannot be ice because ice doesn't exists at a temperature of 100°C.
A. The particles are packed more tightly in materials with more density which causes the vibrations to bounce of the partials more rapidly which makes them go faster
Answer:
3.25 seconds
Explanation:
It is given that,
A person throws a baseball from height of 7 feet with an initial vertical velocity of 50 feet per second. The equation for his motion is as follows :

Where
s is the height in feet
For the given condition, the equation becomes:

When it hits the ground, h = 0
i.e.

It is a quadratic equation, we find the value of t,
t = 3.25 seconds and t = -0.134 s
Neglecting negative value
Hence, for 3.25 seconds the baseball is in the air before it hits the ground.
Answer:
The general equation for conservation of momentum during a collision between n number of objects is given as: [m i ×v i a ] = [m i ×v i b ] Where m i is the mass of object i , v i a is the velocity of object i before the collision, and v i b is the velocity of object i after the collision.
Explanation:
C₂H₃O₂⁻ is an anion.
<u>Explanation:</u>
NaC₂H₃O₂(s) → Na⁺(aq) + C₂H₃O₂⁻(aq)
NaC₂H₃O₂ when dissociated, yields Na⁺ and C₂H₃O₂⁻.
Anion is a negatively charged ion.
In this case, C₂H₃O₂⁻ is an anion.