Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.
- 1. Ball A will have the greater density
- 2. Ball C and Ball D have the same density.
- 3. Ball Q will have the greater density.
- 4. Ball X and Y will have the same density
The density of an object is given as its mass per unit volume of the object.
Mathematically;.
For Case 1:
- Va = Vb and Ma = 2Mb
- D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
- Therefore, the density of ball A,
- D(a) = 2D(b).
- Therefore, ball A has the greater density.
For Case 2:
- D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd
- Therefore, ball C and D have the same density
For Case 3:
- Vp = 2Vq and Mp = Mq
- D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
- Therefore, the density of ball P is half the density of ball Q
- Therefore, ball Q has the greater density.
For case 4:
Therefore, Ball X and Ball Y have the same density.
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Answer:
dt/dx = -0.373702
dt/dy = -1.121107
Explanation:
Given data
T(x, y) = 54/(7 + x² + y²)
to find out
rate of change of temperature with respect to distance
solution
we know function
T(x, y) = 54 /( 7 + x² + y²)
so derivative it x and y direction i.e
dt/dx = -54× 2x / (7 +x² + y²)² .........................1
dt/dy = -54× 2y / (7 + x² + y²)² .........................2
now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2
dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²
dt/dx = -0.373702
and
dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²
dt/dy = -1.121107
During an exothermic reaction; light and heat are released into the environment.
An exothermic reaction is one in which heat is released to the environment. This heat can be physically observed sometimes like in an a combustion reaction.
In an exothermic reaction, the enthalpy of the reactants is greater than the enthalpy of the products.
This heat lost is sometimes felt as the hotness of the vessel in which the reaction has taken place.
In conclusion, light and heat are released into the environment in an exothermic reaction.
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Answer:
Gas
Explanation:
The gas state of matter has the most energy because of how freely the molecules move
To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,
Frequency 
Speed of sound in air 
The definition of wavelength is,
Here,
v = Velocity
f = Frequency
Replacing,
Therefore the wavelength of that tone in air at standard conditions is 0.589m
