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3 years ago

9

Place gamma rays, infrared, microwaves, radio waves, ultraviolet, visible light, and x-rays in order from largest wavelength to

smallest wavelength.

1 answer:

Brums [2.3K]3 years ago

7 0

Answer:

Going by EM SPECTRUM WE HAVE

radio waves, microwaves, infrared, VISIBLE LIGHT, ultraviolet, X-rays, GAMMA RAYS

Explanation:

BECAUSE

V= WAVELENGTH/ FREQUENCY

AS FREQUENCY INCREASES WAVELENGTH DECREASE AN VICE VERSA

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What type of rays would you expect to be used frequently at a hospital to make medical diagnoses?

Bess [88]

X rays because to see your bones

7 0

3 years ago

What can you say about the magnitudes of the forces that the balloons exert on each other?

maxonik [38]

Answer:

$F_G=G. \frac{m_1.m_2}{R^2}$ gravitational force

$F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}$ electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

$F_G=G. \frac{m_1.m_2}{R^2}$

where:

$G=$ gravitational constant $=6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}$

$m_1\ \&\ m_2$ are the masses of individual balloons

$R=$ the radial distance between the center of masses of the balloons.

But when there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

$F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}$

where:

$\rm \epsilon_0= permittivity\ of\ free\ space$

$q_1\ \&\ q_2$ are the charges on the individual balloons

R = radial distance between the charges.

3 0

3 years ago

The fixed hydraulic cylinder C imparts a constant upward velocity v = 2.2 m/s to the collar B, which slips freely on rod OA. Det

Olenka [21]

Answer:

so angular velocity is 7.13128 sec−1

Explanation:

velocity v = 2.2 m/s

displacement s = 220 mm = 0.220 m

distance d = 510 mm = 0.510 m

to find out

angular velocity

solution

we know that

angular velocity will be velocity ( v) / (displacement² + distance²) .....1

now put all these value in equation 1 and we get angular velocity i.e.

angular velocity = velocity ( v) / (displacement² + distance²)

angular velocity = 2.2 / (0.22² + 0.51²)

angular velocity = 2.2 / 0.3085

angular velocity = 7.13128

so angular velocity is 7.13128 sec−1

6 0

3 years ago

Newton’s laws do not apply to small objects?

Soloha48 [4]

Answer:

Yes Newton's laws apply to small objects

EX: Newton s first law

when body at rest always want to be at rest

or body at motion always want to be at motion

unles an external force acts upon it

for example a eraser on the table will be at rest

if so e apply some force then it comes motion

so, Newton s law apply to small object s

7 0

3 years ago

A dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward

marissa [1.9K]

Answer:16.096

Explanation:

Given

mass of dog $\left ( m_d\right )=10kg$

mass of boat $\left ( m_b\right )=46kg$

distance moved by dog relative to ground= $x_d$

distance moved by boat relative to ground= $x_b$

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0= $m_d\times x_d+m_b\dot x_b$

0= $10\times x_d+46\dot x_b$

$x_d=-4.6x_b$

i.e. boat will move opposite to the direction of dog

Now

$|x_d|+|x_b|=7.8$

substituting $x_d$ value

$5.6|x_b|=7.8$

$|x_b|=1.392m$

$|x_d|=6.4032m$

now the dog is 22.5-6.403=16.096m from shore

4 0

3 years ago