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3 years ago

Which type of electromagnetic waves are dangerous enough to be used to kill cancerous cells?

Physics

2 answers:

Ilya [14]3 years ago

8 0

Your answer is A. Gamma Rays

guapka [62]3 years ago

4 0

Answer:

Gamma rays

Explanation:

In the electromagnetic spectrum, the energy of gamma rays is maximum and even their penetrating power is very high.

The increasing order of energy of electromagnetic spectrum is given by

Radio waves < Micro waves < Infrared rays < Visible rays < Ultra violet rays < X rays < Gamma rays

So, gamma rays are highly energetic and they can kill the cancerous cells in the body.

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If a cannonball is fired horizontally from a cannon high atop a mountain, its____?

mel-nik [20]

Acceleration due to gravity will be constant, but the speed can change.

5 0

3 years ago

Read 2 more answers

What is the stopping distance if the car is initially traveling at speed 6.0v? assume that the acceleration due to the braking i

Sunny_sXe [5.5K]

The distance for any rectilinear motion at constant acceleration is:

d = v₀t + 0.5at²
where
v₀ is the initial velocity

So, if v₀ = 6v, and it stopped to 0 m/s, then the acceleration is equal to:
a = (0 - 6v)/t = -6v/t
Thus,
d = (6v)(t) + (0.5)(-6v/t)(t²)
d = 6vt - 3t
<span>d = 3t(2v - 1)</span>

4 0

3 years ago

You are observing a spacecraft moving in a circular orbit of radius 100,000 km around a distant planet. You happen to be located

Natalija [7]

To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:

$v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})$

Here,

$v_{o} =$ Orbital Velocity

$\lambda_{max} =$ Maximal Wavelength

$\bar{\lambda}} =$ Average Wavelength

c = Speed of light

Replacing with our values we have that,

$v_{o} = (3*10^5) (\frac{3.00036-3}{3})$

<em>Note that the average signal is 3.000000m</em>

$v_o = 36 km/s$

Now using the definition about centripetal acceleration we have,

$a_c = \frac{v^2}{r}$

Here,

v = Orbit Velocity

r = Radius of Orbit

Replacing with our values,

$a = \frac{(36km/s)^2}{100000km}$

$a= 0.01296km/s^2$

$a = 12.96m/s^2$

Applying Newton's equation for acceleration due to gravity,

$a =\frac{GM}{r^2}$

Here,

G = Universal gravitational constant

M = Mass of the planet

r = Orbit

The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,

$M = \frac{ar^2}{G}$

$M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}$

$M = 1.943028*10^{27}kg$

Therefore the mass of the planet is $1.943028*10^{27}kg$

7 0

3 years ago

Hi there, it is your average pathetic Junior. Anyways, I need help on 8&9 ASAP, this assignment is beyond overdue but hey wh

zheka24 [161]

Answer:

a) -31.36 m/s

b) 50.176 m

Explanation:

<h2>a) Velocity of the bag</h2>

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

$V_{f}=V_{o} +a.t$ (1)

Where:

$V_{f}$ is the final velocity of the supply bag

$V_{o}=0$ is the initial velocity of the supply bag (we know it is zero because we are told <u>it was "dropped", this means it goes to ground in free fall</u>)

$a=g=-9.8m/s^{2}$ is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)