All categories

2 years ago

Explain why xrays are used to take images of inside the body but UV isnt

Physics

1 answer:

Sidana [21]2 years ago

6 0

Answer:

X-rays go all the way through the body, but ultraviolet rays do not.

Explanation:

An x-ray will show inside the body, but uv light isn't strong enough to go all the way through the body.

You might be interested in

Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul

Gnom [1K]

Answer:

$t=2.5\times 10^{-14}\ s$

Explanation:

We know that charge on electron

$q=1.6\times 10^{-19}\ C$

r= 2 nm

We know that force between two charge given

$F=K\dfrac{Q_1Q_2}{r^2}$

Now by putting the value

$F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}$

$F=5.67\times 10^{-11}\ N$

We know that mass of electron

The mass of electron

$m=9.1\times 10^{-31}\ kg$

F= m a

a= Acceleration of electron

a= F/m

$a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2$

$a=6.2\times 10^{19} m/s^2$

$S=ut+\dfrac{1}{2}at^2$

initial velocity given that zero ,u=0

$20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2$

$t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}$

$t=2.5\times 10^{-14}\ s$

3 0

3 years ago

PLS HELP QUICK TYSM!!!!!!!!!!

maksim [4K]

C. 4 N is the answer if not then i be dumb!!!

8 0

3 years ago

Read 2 more answers

As the mass of an object increases, the weight of the object will ______?

Sergio039 [100]

Answer:

also increase

Explanation:

If the mass increases so will the weight.

4 0

2 years ago

If you need 40.0 Nm of torque in order to loosen a nut on a wn

KonstantinChe [14]

Answer:

0.301 m

Explanation:

Torque = Force × Radius

τ = Fr

40.0 Nm = 133 N × r

r = 0.301 m

The mechanic must apply the force 0.301 m from the nut.

6 0

3 years ago

Calculate the pressure exerted on the heel of a boy’s foot if the boy weighs 80 N and he lands on one heel,which has an area of

cricket20 [7]

Pressure at a given surface is given as ratio of normal force and area

so here force due to heel of the shoes is given as 80 N

and the area of the heel is given as 16 cm^2

so we can say

$P = \frac{F}{A}$

here we have

F = 80 N

$A = 16 cm^2 = 16 * 10^{-4} m^2$

$P = \frac{80}{16 * 10^{-4}}$

$P = 5 * 10^4 N/m^2$

so pressure at the surface due to its heel will be 5 * 10^4 N/m^2

3 0

3 years ago