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pogonyaev
3 years ago
14

Mercury is a liquid metal having a density of 13.6 g/mL. What is the

Physics
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer:

\boxed {\boxed {\sf v \approx 33.088 \ mL}}

Explanation:

The formula for density is:

d= \frac{m}{v}

where <em>m</em> is the mass and <em>v</em> is the volume.

The mass is 0.45 kilograms and the density is 13.6 grams per milliliter. The density is given in grams, so we must convert the mass.

There are 1000 grams in 1 kilogram or \frac{1000 \ g} {1 \ kg}. We can multiply the mass by this ratio.

0.45 \ kg * \frac{1000 \ g} {1 \ kg} = 0.45 * 1000 \ g = 450 \ g

Now we have values for the mass and density:

13.6 \ g/mL =\frac{450 \ g}{ v}

Cross multiply.

\frac {13.6 \ g/mL}{1}  =\frac{450 \ g}{ v}

13.6 \ g/mL * v=450 \ g * 1    \\13.6 \ g/mL * v=450 \ g

We are trying to find the volume, so we must isolate that variable.

13.6 and v are being multiplied. The inverse of multiplication is division. Divide both sides by 13.6

\frac {13.6 \ g/mL*v } { 13.6 \ g/mL} = \frac{ 450 \ g} {13.6 \ g/mL}

v= \frac{ 450 \ g} {13.6 \ g/mL}

The grams will cancel.

v= 33.0882353 \ mL\\v \approx 33.088 \ mL

The volume is about 33.088 milliliters.

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Using the velocity-time graph, the displacement can be calculated by the area under the velocity-time graph. At 3 seconds the total displacement is then equal to (4)(2) + (4 + 2)*1/2 = 11 m. Assuming that the starting point is at x = 0, then the particle at t=3s is at x=11 m.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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The box shown on the rough ramp above is sliding up the ramp.calculate the force of friction on the box
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We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:

F_f=\mu N

Where.

\begin{gathered} N=\text{ normal force} \\ \mu=\text{ coefficient of friction} \end{gathered}

To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:

In the diagram we have:

\begin{gathered} m=\text{ mass}_{} \\ g=\text{ acceleration of gravity} \\ mg=\text{ weight} \\ mg_y=y-\text{component of the weight. } \end{gathered}

Adding the forces in the y-direction we get:

\Sigma F_y=N-mg_y

Since there is no movement in the y-direction the sum of forces must be equal to zero:

N-mg_y=0

Now we solve for the normal force:

N=mg_y

To determine the y-component of the weight we will use the trigonometric function cosine:

\cos 40=\frac{mg_y}{mg}

Now we multiply both sides by "mg":

mg\cos 40=mg_y

Now we substitute this value in the expression for the normal force:

N=mg\cos 40

Now we substitute this in the expression for the friction force:

F_f=\mu mg\cos 40

Now we substitute the given values:

F_f=(0.2)(10\operatorname{kg})(9.8\frac{m}{s^2})\cos 40

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F_f=15.01N

Therefore, the force of friction is 15.01 Newtons.

3 0
1 year ago
Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t
elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

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If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

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s = ut + 1/2 at²

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As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only  horizontal velocity initially , Vy = 0

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t = \sqrt{\frac{2h}{g} }

Horizontal distance covered  = t x Vx

= \sqrt{\frac{2h}{g} } \times  V_x

From this expression also

Horizontal distance covered is proportional to \sqrt{h} .

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