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3 years ago

It took 3 seconds for an object that was thrown up with velocity vo from

Physics

1 answer:

otez555 [7]3 years ago

5 0

Answer:

ik u got it bc this was 2 weeks ago

Explanation:

but yes and yuea

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Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500ºC

scoray [572]

Answer:

m = 4.65 kg

Explanation:

As we know that the mass of the water that evaporated out is given as

$m = 0.0250 kg$

so the energy released in form of vapor is given as

$Q = mL$

$Q = (0.0250)(2.25 \times 10^6)$

$Q = 56511 J$

now the heat required by remaining water to bring it from 15 degree to 100 degree

$Q_2 = ms\Delta T$

$Q_2 = (4 - 0.025)(4186)(100 - 15)$

$Q_2 = 1.41\times 10^6J$

total heat required for above conversion

$Q = 56511 + 1.41 \times 10^6 = 1.47 \times 10^6 J$

now by heat energy balance

heat given by granite = heat absorbed by water

$m(790)(500 - 100) = 1.47 \times 10^6$

$m = 4.65 kg$

4 0

4 years ago

How is a pareto chart different from a standard vertical bar graph?

Alona [7]

Answer:

A Pareto chart, named after an Italian economist, combines a bar chart with a line graph. The bar chart is different from a histogram in more than one way. For example, the vertical bars need not touch one another as per a histogram

Explanation:

5 0

4 years ago

Eyeglasses bend light help the image form properly on the retina. The bending of light as it passes from one medium to another i

garri49 [273]

It's refraction....When light bends on passing from one medium to another is refraction !

3 0

4 years ago

Read 2 more answers

A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

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

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0

3 years ago

Course Task

kaheart [24]

Answer:

gravity effects only the mass of an object

Explanation:

an acorn is heavier that a stick and it's round shape gives it less wind resistance so gravity pulls it down faster. A stick vertical has little wind resistance but when rotated to when it is horizontal it has more resistance so it slows down but the the force of gravity is still as strong. A leaf has very little mass and it's shape provideds lots of wind resistance so it falls slower. The moon is not only effected by the earth's gravity it is also effected by the suns. It has its own gravity as well. with all these factors it stays in place and moves with the earth.

4 0

3 years ago