A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

Speed of 1 Kg mass post-collision =?

Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

Response:R=1607556m

θ=180degrees

Clarification:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Utilizing an analytical approach:

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

θ=arctan(-73/160755.9)

θ=-7.9256*10^-6

It is worth noting that since θ is in the second quadrant, 180 is added

θ=180-7.9256*10^6=180degrees

#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

Therefore, its weight in air is noted as

Next, the buoyant force acting on the lead is defined as

We know that

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

#2

The buoyant force acting on the lead block counterbalances its weight

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.

#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

Answer:

The outcome of adding 999mm to 100m is 101m.

Explanation:

That's my belief.