Can u explain cause i dont understand

Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the ________.

Black_prince [1.1K]

Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the collision model.

-Hope this helps.

7 0

3 years ago

Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we

sukhopar [10]

The total momentum of the system is equal to 50 Kgm/s.

Given the following data:

Mass 1 = 80 kg

Velocity 1 = 2 m/s east.

Mass 2 = 70 kg

Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

$Momentum = mass \times velocity$

For Football player 1:

$Momentum = 80 \times 2$

Momentum 1 = 160 Kgm/s.

For Football player 2:

$Momentum = 70 \times 3$

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

$Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160$

Total momentum = 50 Kgm/s.

Note: We subtracted because the football players were moving in opposite directions.

6 0

2 years ago

The weather conditions

bixtya [17]

Answer:

ok confusion but we could figure it out right

Explanation:

<h3>dhdjhdndnd but its fine how was your day tho </h3>

4 0

3 years ago

If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a result?

andrew11 [14]

Answer:

energy of motion decrease

Explanation:

yes

7 0

2 years ago

1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it

spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

PE = (2.5)(98)(14) and

PE = 340 J

b. $KE=\frac{1}{2}mv^2$ so

$KE=\frac{1}{2}(2.5)(14)^2$ and

KE = 250 J

c. TE = KE + PE so

TE = 340 + 250 and

TE = 590 J

d. PE at 8.7 m:

PE = (2.5)(9.8)(8.7) and

PE = 210 J

e. The KE at the same height:

TE = KE + PE and

590 = KE + 210 so

KE = 380 J

f. The velocity at that height:

$380=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(380)}{2.5} }$ so

v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

590 = KE + PE and

PE = (2.5)(9.8)(11.6) so

PE = 280 then

590 = KE + 280 so

KE = 310 then

$310=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(310)}{2.5} }$ so

v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

$v=v_0+at$ and

26 = 0 + 9.8t and

26 = 9.8t so the time at 26 m/s is

t = 2.7 seconds. Now we use that in the equation for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in the time the object was at 26 m/s:

Δx = 0t + $\frac{1}{2}(-9.8)2.7)^2$ so

Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0

3 years ago