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3 years ago

How does physical change affect the identity of a substance

1 answer:

3 0

A physical change affects one or more physical properties of a substance without changing the identity of the substance. No matter how small the pieces of chalk are, each piece still has the same chemical properties. Examples of physical changesinclude boiling water, sanding a piece of wood, and mixing sand and water.

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Within a vacuum, the property to all electromagnetic waves is their

bezimeni [28]

The answer is D) Velocity

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4 years ago

A flea jumps straight up to a maximum height of 0.540 m . what is its initial velocity v0 as it leaves the ground?

jok3333 [9.3K]

For an uniformly accelerated motion, we can write
$2ah=v_f^2-v_0^2$
where $a=g=-9.81 m/s^2$ is the acceleration of this motion, which in this problem is the gravitational acceleration, with a negative sign because it points downward, against the direction of the motion; h=0.540 m is the distance covered by the flea, and $v_0$ is the initial velocity.

At the maximum height, the velocity is zero, so $v_f =0$ . Therefore we can solve to find $v_0$ :
$v_0 = \sqrt{2ah}= \sqrt{2(9.81 m/s^2)(0.540 m)} =3.25 m/s$

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3 years ago

Steel is an alloy of _____. One of the main reasons steel is used more widely than iron is because it’s_____ than iron. thanks f

yarga [219]

1. Iron I think and 2. stronger

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3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced: