The density of a substance equals its mass

A 120-meter-long ski ift carries skiers from a station at the foot of a slope to a second station 40 m above. what is the IMA (i

Pepsi [2]

The ideal mechanical advantage (IMA) of an inclined plane is given by:
$IMA = \frac{L}{h}$
where L is the length of the inclined plane while h is the height.
In our problem, L=120 m and h=40 m. Therefore, the IMA of the ski lift is
$IMA= \frac{120 m}{40 m}=3$

7 0

4 years ago

Where is the epicenter (it's about earthquakes)

sweet [91]

Answer:

The epicenter is the point on the earth's surface vertically above the hypocenter (or focus), point in the crust where a seismic rupture begins.

<h2><u>PLEASE MARK ME BRAINLIEST, PLEASE</u></h2>

4 0

3 years ago

Read 2 more answers

A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.

Contact [7]

Answer:

A) $V_A = 11.93~V$

B) The vector definition of E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

$V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}$

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

$r = \sqrt{x^2+4^2}$

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

$dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}$

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

$V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx$

By using an integral table, this can be calculated:

$V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V$

B) The electric field can be found by a similar approach, but a different formula:

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

Let's apply this formula to the infinitesimal part we have chosen.

$dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)$

By the geometry sine and cosine terms can be found:

$\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}$

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

$E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx = 2.41(\^y)$

So, the final E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

The magnitude of the E-field is

E = 2.66 N/m

6 0

3 years ago

What is the relationship between the size of a predator population and the size of a prey population

Contact [7]

<span>Let's put it this way. Say you have a killer-whale and a penguin. Killer-whales are major predators to penguins. Now, say the killer-whale population increases. The penguins would be eaten more by the killer-whales, then causing a population decrease for the penguins. If the population decreases, they're won't be enough penguins, and they most likely will become extinct, as well as causing a population decrease for the killer-whales as well. Whereas, vis versa, they're were a killer-whale population decrease. The penguins would be less hunted, therefore, creating a population increase for the penguins.</span>

3 0

4 years ago

A body of mass m, = 0.050 kg and initial velocity v₁ = 9 m / s turned to the right collides with a body of mass m₂ = 0.030 g wit

Andrej [43]

Answer:

1 m/s

Explanation:

Po = Pf

(0.050kg)(9 m/s) + (0.030kg)(v2) = (0.05kg + 0.030)(6 m/s)

0.45 kgm/s + (0.030kg)(v2) = 0.48 kgm/s

(0.030kg)(v2) = 0.03 kgm/s

v2 = 1 m/s

4 0

2 years ago