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Kryger [21]
3 years ago
8

A block of wood is 4 cm wide, 5 cm long, and 10 cm high. It weighs 100 grams. Calculate its volume. Calculate its density. Will

the block sink or float in water? Explain. (Remember, the density of water is 1.0 g/cm3.)
Physics
1 answer:
77julia77 [94]3 years ago
3 0

Answer:

V=200cm^3\\\\\rho =0.500g/cm^3

It will float.

Explanation:

Hello.

In this case, given the width, length and height, we can compute the volume as follows:

V=W*L*H\\\\V=4cm*5cm*10cm\\\\V=200cm^3

Moreover, since the density is computed via the division of the mass by the volume:

\rho =\frac{m}{V}

We obtain:

\rho =\frac{100g}{200cm^3} \\\\\rho =0.500g/cm^3

In such a way, since the solid has a lower density than the water, we infer it will float.

Best regards.

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Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of dis
Sedaia [141]

Answer: Please find the answer in the explanation

Explanation:

Under what circumstances does distance traveled equal magnitude of displacement?

When a body's motion is linear in one direction. Or a body moving in a straight line without turning back.

What is the only case in which magnitude of displacement and distance are exactly the same?

When the body is moving in a straight line with without changing direction or without turning back.

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3 years ago
A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius
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Answer:

<u>r < a:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}

<u>r = a:</u>

E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}

<u>a < r < b:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

<u>r = b:</u>

E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}

<u>b < r < c:</u>

E = 0

<u>r = c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}

<u>r < c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

Explanation:

Gauss' Law will be applied to each region to find the E-field.

\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

<u>r<a:</u>

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}

Applying Gauss' Law:

E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}

The minus sign determines the direction of the field, which is towards the center.

<u>At r = a: </u>

E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}

<u>At a < r < b:</u>

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u />E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}<u />

<u>At r = b:</u>

<u />E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}<u />

Again, the minus sign indicates the direction of the field towards the center.

<u>At b < r < c:</u>

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

<u>At r < c:</u>

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

<u>At r = c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}

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Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.

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