Can you get cool powers.

The _______ are found on the right side of the arrow in a chemical reaction.

Dimas [21]

Reactants ---> product
The products are found on the right side of the arrow in a chemical reaction.

5 0

3 years ago

Two points A=(+16) and B=(-4) At electric potential respectively find:

Damm [24]

Answer:

a. Point A

b. 20 V

c. 100 J

Explanation:

a. Point A is at a higher potential because there is a positive sign in front of its magnitude. Since it is a positive integral value, and has a higher magnitude than point B which is at -4, point A is thus at a higher potential than point B.

b. The potential difference between the two points ΔV = A - B

= +16 V - (-4 V)

= +16 V + 4 V

= + 20 V

c. The work done, W in moving a charge Q across a potential difference ΔV is W = QΔV

So, since Q = 5 C and ΔV = + 20 V

W = QΔV

= 5 C × (+ 20 V)

= 100 J

4 0

3 years ago

The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic

Drupady [299]

a) Density at 100 degrees: $1.34\cdot 10^4 kg/m^3$

Explanation:

The density of mercury at 0 degrees is $d=1.36\cdot 10^4 kg/m^3$

Let's take 1 kg of mercury. Its volume at 0 degrees is

$V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3$

The formula to calculate the volumetric expansion of the mercury is:

$\Delta V= \alpha V \Delta T$

where

$\alpha=180\cdot 10^{-6} K^{-1}$ is the cubic expansivity of mercury

V is the initial volume

$\Delta T$ is the increase in temperature

In this part of the problem, $\Delta T=100 C-0 C=100 C=100 K$

So, the expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3$

b) Density at 22 degrees: $1.355\cdot 10^4 kg/m^3$

We can apply the same formula we used before, the only difference here is that the increase in temperature is

$\Delta T=22 C-0 C=22 C=22 K$

And the volumetric expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3$

8 0

3 years ago

________ reaction time involves selecting a specific and correct response from several choices when presented with several diffe

notsponge [240]

The answer would be complex reaction time

4 0

3 years ago

Choose the +x-direction to point to the right. • Object 1 has a mass 1.66 kg and is moving to the right at 11.2 m/s. • Object 2

egoroff_w [7]

Answer:

M = 49.4kgm/s (towards the left)

Explanation:

Momentum is the product of mass and velocity of an object

Momentum = mass * velocity

Momentum of Object 1 with mass 1.66 kg moving to the right at 11.2 m/s, is expressed as:

M1 = 1.66 * 11.2

M1 = 18.592kgm/s

Momentum of Object 2 with mass 6.59 kg moving to the left at 10.4 m/s is expressed as:

M2 = 6.59 * -10.4

M2 = -68.536kgm/s (negative since it is moving towards the left)

The total momentum will be the sum of momentum along the x-component as shown:

M = M1+M2

M = 18.592kgm/s--68.536kgm/s

M = -49.944kgm/s

M = 49.4kgm/s (towards the left)

5 0

3 years ago