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2 years ago

A measure of change in velocity in a measure of time is

1 answer:

8 0

$\\ \sf\Rrightarrow Acceleration=\dfrac{\Delta V}{t}[)tex]In simplified form[tex]\\ \sf\Rrightarrow Acceleration=\dfrac{dv}{dt}$

The rate of change of velocity with respect to time period is called acceleration.
SI units-m/s^2

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3 kg of wet clothes are hung on the middle of a clothesline with posts 6 ft apart. The clothesline sags down by 3 feet. What is

miskamm [114]

Answer:

Tension in string equals 14.715 Newtons

Explanation:

The situation is represented in the figure attached below:

For equilibrium of the clothes along y- axis we have

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{W}{2cos(\theta )}$

Applying values we get

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{3\times 9.81}{2\times 1}=14.715N(\because cos(\theta )=\frac{3}{3}=1)$

4 0

2 years ago

Read 2 more answers

An object weighing 2.7n while in air and 1.2n when completely immersed in water.find the relative density of the object

Elden [556K]

Answer:

1.8 = relative density (there are no units for relative density)

Explanation:

It displaces water equal to it's volume and gets buoyancy equal to that amount of water

2.7 - 1.2 = 1.5 N of buoyancy

density of water = 1 gm /cc

1.5 N = m (9.81)

m of water displaced = .1529 kg

152.9 cc of water will produce this buoyancy....this is the volume of the object

find mass of object 2.7 = m (9.81) shows m = .2752 kg = 272.5 gm

density = mass/ volume = 272.5 / 152.9 = 1.8 gm / cc

Relative to water (which is 1 gm / cc) the relative density is 1.8

====> it is 1.8 times denser than water and will sink when in water....

8 0

2 years ago

Clouds of gas and dust as well as new star formation are typically seen in __________galaxy

ohaa [14]

Answer:

Clouds of gas and dust as well as new star formation are typically seen in Spiral galaxy

Explanation:

The ingredients for star formation are gas and dust, a spiral galaxy is a group of stars that are together as a consequence of the gravitational pull between them and, as the name said it, has a spiral shape.

They are composed for a central bulge, a disk (in which a spiral arms structure can be seen), gas and dust. Because of the presence of gas and dust, star formation is possible.

3 0

2 years ago

A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radi

svetlana [45]

Answer:

$E=0$ at r < R;

$E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}$ at 2R > r > R;

$E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}$ at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

$\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon}$ (integral over a closed surface)

where,

$E$ = Electric field

$Q_{enclosed}$ = charged enclosed within the closed surface

$\epsilon$ = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

$Q_{enclosed}$ = 0 and hence $E$ = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

$Q_{enclosed}$ = Q,

therefore,

$E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}$

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to $4\pi r^{2}$ )

or, $E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}$

at r >= 2R

$Q_{enclosed}$ = 2Q

Hence, by similar calculations, we get,

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

4 0

3 years ago

A simple ideal Brayton cycle uses argon as the working fluid. At the beginning of the compression, P1 = 15 psia and T1 = 70°F, t

Shtirlitz [24]

B I thing hope this helps

7 0

3 years ago