Which of the following is a mixture? A. steel B. water c. oxygen D. gold

If you stand on a bathroom scale,the spring inside the scale compresses 0.50 mm, and it tells you your weight is 700 N. Now if y

Tanzania [10]

Initially, mg = kx. K = mg/x = 700/0.5x10^-3 = 1400000N/m. From second condition, applying work-energy theorem, potential enery- elastic potential energy = change in kinetic energy. Now change in kinetic energy is 0 since initial and final velocities are 0m/s. Therefore, potential energy = elastic potential energy. mgh = (1/2) * k* x^2. x^2 = 2(mg)h/k = 2 x 700 x 1.3/ 1400000. x = 0.036m. Hope it's clear.

3 0

4 years ago

Why do the giant planets and their moons have compositions different from those of the terrestrial planets?

dangina [55]

Answer and Explanation:

The formation of planets ,initially was the result of gradual accumulation of solid matter into the solar nebula. As a result of high temperature in the interior of our solar system, metals and rocks were the only materials to get compressed.

The matter that was volatile could not be compressed so close to the heat energy radiated by the early Sun.

On the outer part of the solar system, solid matter included hydrogen compounds, rocks and metals with a lot of matter for planet formation.

The Giant planets were formed by capturing Helium and hydrogen gases as well whereas the terrestrial planets being much more smaller are made up of mainly rocks like silicates and metals like iron.

The moons of terrestrial planets like that of Earth is also terrestrial in nature consisting of rocks and metals as the constituent material while that of giant planets consist of frozen water in half the proportion and the other half is rocks and metals.

4 0

4 years ago

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

3 years ago

An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down

soldier1979 [14.2K]

Answer:

$1.9\times 10^{-4}$

$1.2\times 10^{-4}\ m$

Explanation:

r = Radius = 2.7 cm

F = Force = $3.2\times 10^4\ N$

A = Area = $\pi r^2$

$\sigma$ = Stress = $\frac{F}{A}$

E = Young's modulus = $7\times 10^{10}\ Pa$

$\epsilon$ = Strain

$L_0$ = Original length = 67 cm

$\Delta L$ = Change in length

Young's modulus is given by

$E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}$

Strain is $1.9\times 10^{-4}$

Strain is given by

$\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m$