All categories

4 years ago

What is an example of radiation? check all that apply

1 answer:

4 0

There are no appropriate examples in the list you provided with your question.

Examples of radiation:

... sunshine to tan your skin
... radio energy to bring you the news
... X-ray to check your teeth
... microwave to heat up the meatloaf
... flashlight to see where you're going
... RF energy to get an MRI of your knee
... infrared radiation from the campfire to warm your tootsies
... UHF radio waves to make a call or check Facebook with your smartphone

You might be interested in

So we were working on some pulley problems but this one has kinda left me scratching my head, please help! My question is for pa

USPshnik [31]

Explanation:

Sum of forces on B in the -y direction:

∑F = ma

mBg − T = mBa

Sum of forces on A in the +x direction:

∑F = ma

T = mAa

Substitute:

mBg − mAa = mBa

mBg − mBa = mAa

mA = mB (g − a) / a

Plug in values:

mA = (5 kg) (10 m/s² − 0.01 (10 m/s²)) / (0.01 (10 m/s²))

mA = 495 kg

The answer key seems to have a mistake. It's possible they meant mB = 1 kg, or they changed mB to 5 kg but forgot to change the answer.

8 0

3 years ago

Traveler A starts from rest at a constant acceleration of 6 m/s^2. Two seconds later, traveler B starts with an initial velocity

Troyanec [42]

Answer:

3. 3.5 s

Explanation:

The position of traveller A is given by the equation:

$x_A(t) = \frac{1}{2}a t^2$

where

$a = 6 m/s^2$ is the acceleration of A

t is the time measured from when A started the motion

The position of traveller B instead is given by

$x_B(t) = u_B (t-2) + \frac{1}{2}a(t-2)^2$

where a (acceleration) is the same as traveller A, and

$u_B = 20 m/s$

is B's initial velocity. We can verify that the formula is correct by substituting t=2, and we get $x_B=0$ , which means that B starts its motion 2 seconds later.

Traveller B overtakes traveller A when the two positions are the same, so:

$x_A = x_B\\\frac{1}{2}at^2 = u_B (t-2) + \frac{1}{2}a(t-2)^2\\\frac{1}{2}at^2 = u_B t - 2u_B +\frac{1}{2}at^2 +2a-2at\\u_Bt-2at = 2u_B-2a\\t=\frac{2u_B-2a}{u_B-2a}=\frac{2(20)-2(6)}{20-2(6)}=3.5 s$

4 0

3 years ago

A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0

n200080 [17]

Answer:

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

Work done = $m*g*h$

where m is mass

g is acceleration due to gravity

and h is the height

Work done in terms of force and distance is = mg

where m is mass of bucket and

g is acceleration due to gravity

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = $\Delta P.E$

W = m*g*h

where g = 9.9 m/ $s^2$

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

6 0

3 years ago

A monochromatic light beam is incident on a barium target that has a work function of 2.50 \mathrm{eV} . If a potential differen

leva [86]

The wavelength of the light beam required to turn back all the ejected electrons is 497 nm which is option (b).

Work function is a material property defined as the minimum amount of energy required to infinitely remove electrons from the surface of a particular solid.
The potential difference required to support all emitted electrons is called the stopping potential which is given by $v_0=\frac{K.E_m_a_x}{e}$ .....(1)
where $v_0$ is the stopping potential and e is the charge of the electron given by $1.6\times10^-^1^9$ .

It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation is given by:

$K.E_m_a_x=E-\phi$ ....(2)

where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

$ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV$

As we know that $E=\frac{hc}{\lambda}$ ....(3)

where Speed of light, $c = 3\times10^8 m/s$ and Planck's constant , $h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs$

From equation (3) , we get

$\lambda=\frac{hc}{E} \\\\\lambda=\frac{ 4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm$

Learn about more einstein photoelectric equation here:

brainly.com/question/11683155

#SPJ4

8 0

1 year ago

Which sector is not part of the circular flow model of the economy ?

morpeh [17]

Answer:

a. government sector

Explanation:

7 0

3 years ago

Read 2 more answers