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3 years ago

If all the stars in an elliptical galaxy traveled random directions in their orbits, the elliptical galaxy would be type

1 answer:

7 0

The answer would be E7. Galaxies categorized as E0 look to be nearly perfect, while those registered as E7 seem much extended than they are widespread. It is worth noting, though, that a galaxy's look is connected to how it lies on the sky when viewed from Earth. An E7 galaxy is very long and thin or the flattest of them all.

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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

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

The cylinder height decreases by $1.2\times 10^{-4}\ m$

3 0

3 years ago

Name each type of symbiosis and explain how the two species are affected

disa [49]

Frogs
snakes if there food chain is mesesed up it dont work no more

7 0

3 years ago

A hockey player uses her hockey stick to exert a force of 6.81 N on a stationary hockey puck. The hockey puck has a mass of 165

Anna007 [38]

Answer:

41.3 m/s^2 option (e)

Explanation:

force, F = 6.81 N

mass, m = 165 g = 0.165 kg

Let a be the acceleration of the puck.

Use newtons' second law

Force = mass x acceleration

6.81 = 0.165 x a

a = 41.27 m/s^2

a = 41.3 m/s^2

Thus, the acceleration of the puck is 41.3 m/s^2.

5 0

3 years ago

what is the approximate weight of a 20-kg cannonball on the moon if the acceleration due to gravity is 1.6m/s^2

monitta

On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons

BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N

5 0

3 years ago

A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.

Anastasy [175]

Initially, the velocity vector is $\langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle$ . At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by $4.9(0.2)^2$ , so the velocity is $\langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle$ .

Converting back to direction and magnitude, we get $\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle$

4 0

3 years ago