Anybody know this answer?

Usually, when the temperature is increased, what will happen to the rate of dissolving?

ArbitrLikvidat [17]

It will decrease

When the temperature increased, the rate will decrease

5 0

3 years ago

When you measure the boiling point of mercury, you are investigating a ___. a.chemical change b.chemical property c.physical cha

Amanda [17]

D. physical property

the bonds between molecules of mercury are breaking so it's physical and it's not changing the chemical composition of the substance

3 0

2 years ago

With fuel prices for combustible engine automobiles increasing, researchers and manufacturers have given more attention to the c

Alinara [238K]

Answer:

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

Explanation:

An object's weigh due to the gravitational attraction force of the earth is:

w = mg

Where: m is the object's mass

g is the gravitational acceleration in the surface earth

g = 9.8 m/s2

The the ultralight car's weight is:

$w_{uc} = (540)(9.8)$

$w_{uc} = 5292 N$

And the Honda Accord's weight is:

$w_{HA} = (1450)(9.8)$

$w_{HA} = 14210 N$

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

4 0

3 years ago

Why do birds not get shock when they<br>sit on high power live wire but we do?<br>

Marina86 [1]

Answer:

Their bodies don't conduct electricity like we do.

Explanation:

6 0

2 years ago

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A skydiver jumps out of a hovering helicopter. A few seconds later, another diver jumps out, so they both fall along the same ve

Sergio039 [100]

Answer:

distance difference would a) increase

speed difference would f) stay the same

Explanation:

Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

Their equations of motion for distance and velocities are

$s_2 = gt^2/2$

$s_1 = g(t + \Delta t)^2/2$

$v_2 = gt$

$v_1 = g(t + \Delta t)$

Their difference in distance are therefore:

$\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2$

$\Delta s = g/2((t + \Delta t)^2 - t^2)$

$\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t)$ (As $A^2 - B^2 = (A-B)(A+B)$

$\Delta s = g\Delta t/2(2t + \Delta t)$

So as time progress t increases, Δs would also increases, their distance becomes wider with time.

Similarly for their velocity difference

$\Delta v = v_1 - v_2 = g(t + \Delta t) - gt$

$\Delta v = gt + g\Delta t - gt = g\Delta t$

Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.

This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

8 0

2 years ago

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