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3 years ago

Please help me on questions 3 and 6, thank you! :D I'll give brainliest!

Physics

2 answers:

KIM [24]3 years ago

6 0

3. To increase gravitational pull to the earth, the more mass an object has, the stronger the gravitational pull, so increase mass.

6. While sound energy depends on a medium, radiant energy uses it's own sources to stay independent from other mediums.

Anarel [89]3 years ago

4 0

Answer:

3. if you increase your mass you also increase the gravitational pull

6. Radiant energy doesn't depend on a medium and sound energy is dependent on a medium.

Explanation:

i hope this helps-

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Find the change in velocity of a 369 g hockey puck subject to the force shown below.

Elena-2011 [213]

Answer:

The change in velocity is 15.83 [m/s]

Explanation:

Using the Newton's second law we have:

ΣF = m*a

The force in the graph is 185 N, therefore:

$185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]$

$a=501.35[m/s^2]$

Now using the following kinematic equation:

$V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m\\$

Now replacing the values:

$V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]$

3 0

3 years ago

The fumes of lighted 'incense stick' spread in the whole room due to-

Firlakuza [10]

Answer: burning of the incense and air drift in the room

Explanation: the smoke will automatically rise to the top of the room and then with air drift if would help spread further around the room and spread the smell

7 0

3 years ago

A sample of aluminum has a mass of 4.86 grams and a volume of 1.8 cubic centimeters. What is the density of aluminum rounded to

TEA [102]

Answer:

3g/cm³

Explanation:

2.7

The seven adds 1 to the 2 making it 3.0

6 0

3 years ago

The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago

A planet orbits a star in an elliptical orbit. At a particular instant the momentum of the planet is

Masja [62]

Answer:

L = m v r (The momentum remains constant)

Explanation:

Even in an ellipsoidal orbit, the law of conservation of angular momentum always apply. When the plant approached the perihelion, the radius of the orbit decreases and the speed of the star increases to conserve the momentum. Similarly, when the planet approaches the aphelion, the speed of the star decreases as the radius increases to conserve the momentum. So, the momentum at a particular instant can be calculated by L = m v r

7 0

3 years ago