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4 years ago

What kind of relationship benefits both parties involved?

Physics

2 answers:

diamong [38]4 years ago

7 0

Mutualism benefits both parties

harina [27]4 years ago

5 0

The heart of "mutualism" is that two species live together in harmony. Both species receive an advantage<span> by working with the other. </span>

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A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph

arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

$I_{CM} = \dfrac{2}{5}mR^2$ .....(1)

now, moment of inertia about tangent

$I= \dfrac{2}{5}mR^2 + mR^2$

$I= \dfrac{7}{5}mR^2$ ...........(2)

dividing equation (1)/(2)

$\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}$

$\dfrac{I_{CM}}{I}=\dfrac{2}{7}$

$I_{CM}=\dfrac{2}{7}I$

the correct answer is option E

4 0

3 years ago

What is the maximum kinetic energy of electrons ejected from barium when illuminated by white light, to 750 nm?

Ugo [173]

Answer:

$1.0045\times10^{-19}$ J

Explanation:

The complete statement is given as

What is the maximum kinetic energy of electrons ejected from barium (Wo = 2.48 eV) when illuminated by white light, 400 nm to 750 nm ?

$\lambda$ = wavelength of the white light = 400 nm = 400 x 10⁻⁹ m

$c$ = speed of white light = 3 x 10⁸ ms⁻¹

Energy of the white light is given as

$E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{400\times10^{-9}}\\E = 4.9725\times10^{-19} J$

$W_{o}$ = Work function of barium = 2.48 eV = 2.48 x 1.6 x 10⁻¹⁹ J

$K$ = Maximum kinetic energy

Using photoelectric equation we have

$E = W_{o} + K\\4.9725\times10^{-19} = 2.48\times1.6\times10^{-19} + K\\K = 1.0045\times10^{-19} J$

3 0

3 years ago

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

padilas [110]

Answer:

Explanation:

First case

A cart speed is 0.3m/s. i.e the initial velocity is u=0.3m/s

It collide with a stationary body, then after collision the ball rebounds and move in opposite direction. This shows that the ball have a velocity after impulse let say v

Then, impulse is given as the change in linear momentum of a body

Impulse =m∆v

I=m(v-u)

Note, momentum is a vector quantity.

I=m(v--u)

I=m(v+u)

I=m(v+0.3)

I¹=0.3m+mv. Equation 1

Second case

A cart speed is 0.3m/s. i.e the initial velocity is u=0.3m/s

It collide with a stationary body, then after collision the ball is at rest, this show that the final velocity is v=0

Then, impulse is given as the change in linear momentum of a body

Impulse =m∆v

I=m(v-u)

Note, momentum is a vector quantity.

I=m(v--u)

I=m(v+u)

In this case v=0 u=0.3m/s

I=m(0+0.3)

I²=0.3m. Equation 2

If we compare impulse 1 (I¹) to impulse 2 (I²)

Subtract equation 2 from 1

We have, I¹ - I² =0.3m+mv -0.3m

I¹ - I² =mv

I¹ =mv+I²

We notice that the first impulse (I¹) is greater than second impulse (I²) by mv.

The correct answer is A

5 0

3 years ago

In the first video for chapter 29, we looked at the emission spectrum of excited gases. In this video, we look at the __________

Elan Coil [88]

Although the video is not found here, the sentence makes reference to the transmission spectrum of colored filters.

<h3>What is the transmission spectrum?</h3>

The transmission spectrum indicates the light portion having a given wavelength that can be passed through a filter.

This spectrum (transmission spectrum) depends on the physical separation of the particles that form the filter.

In conclusion, although the video is not found here, the sentence makes reference to the transmission spectrum of colored filters.

Learn more about the transmission spectrum here:

brainly.com/question/1287536

#SPJ1

6 0

3 years ago

A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8

mylen [45]

Answer:

$f = 2.6 \times 10^{-13}$

Explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

$N = \frac{m}{29} \times 6.02 \times 10^{23}$

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

$N_e = m(6.02 \times 10^{29})$

now diameter of the ball is 7.0 mm

density of the ball = $8900 kg/m^3$

now we have

$m = (\frac{4}{3}\pi r^3)(8900)$

$m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)$

$m = 1.6 gram$

now we have

$N_e = 9.63 \times 10^{23}$

now the charge on the copper ball is 40 nC

so the number of electrons removed

$Q = ne$

$40 \times 10^{-9} = n(1.6 \times 10^{-19}$

$n = 2.5 \times 10^{11}$

so the fraction of number of electrons removed is given as

$f = \frac{n}{N_e}$

$f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}$

$f = 2.6 \times 10^{-13}$

7 0

4 years ago