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4 years ago

How does radiation help treat cancer?

Physics

2 answers:

Alex Ar [27]4 years ago

6 0

Answer:

Explanation:

sergejj [24]4 years ago

3 0

Answer:

it cools down the cancer cells

Explanation:

The cancer cells are rapidly reproducing and when you cool it down it slows down the mutation/ reproduction process.

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A projectile is launched at an angle 60° to the horizontal

Katena32 [7]

Explanation:

We have,

A projectile is launched at an angle 60° to the horizontal with some initial speed 45 m/s.

The cliff is 265 m high.

(A) The final velocity of the projectile is given by :

$v_f^2-v_i^2=2gs\\\\v_f^2=2gs+v_i^2\\\\v_f^2=2\times 9.8\times 265+45^2\\\\v_f=84.96\ m/s$

(B) The maximum height of the projectile is given by :

$H=\dfrac{v_i^2\sin^2\theta}{2g}\\\\H=\dfrac{45^2\times \sin^2(60)}{2\times 9.8}\\\\H=77.48\ m$

$R=\dfrac{v_i^2\sin 2\theta}{g}\\\\R=\dfrac{(45)^2\times \sin 2(60)}{9.8}\\\\R=178.94\ m$

7 0

4 years ago

How is it possible for one electrically neutral atom to exert an electrostatic force on other electrically neutral atom?

Free_Kalibri [48]

Answer:

Explained

Explanation:

Although Atom are electrically neutral. But atom atom is combination of nucleus and electrons. The nucleus of the atom is composed neutron and positively charged protons. On the outside of nucleus at some distance are the electrons which are negatively charged. So, there is difference in position of the two differently charge species. So, this way a electrically neutral atom can exert a electrostatic force on other electrically neutral atom

8 0

4 years ago

Consider a system made of a ramp that has a run of 8 m and a rise of 6 m. Two equal masses of 4-kg are connected to each other b

djyliett [7]

Answer:

The force of friction is 15.68 N.

Explanation:

Given:

Rise is 6 m and run is 8 m.

Mass of each body (m) = 4 kg

The whole system is in equilibrium.

Now, consider the diagram below representing the scenario given in the question.

The forces acting on the mass that is hanging are tension force in the upward direction and weight of the body acting vertically downward.

As the mass is in equilibrium, the total upward force equals total downward force. So,

$T= mg=4\times 9.8=39.2\ N$ -------- (1)

Now, the forces acting on the other mass along the ramp are:

1. Tension (T) up the ramp

2. $mg sin\theta$ and frictional force (f) down the ramp

Now, as per question:

Rise = 6 m and run = 8 m

So, from figure,

$\tan\theta=\frac{6}{8}=0.75\\\\\theta=\tan^{-1}(0.75)=37\°$

Now, $\sin\theta=\sin(37\°)=0.6$

Now, as the other mass is also at rest, net force acting on it is also 0. So,

$F_{net}=0\\\\T-(mg\sin\theta+f)=0\\\\mg\sin\theta+f=T\\\\f=T-mg\sin\theta$

Now, plug in the given values and solve for 'f'. This gives,

$f=39.2-(4\times 9.8\times 0.6)\\\\f=39.2-23.52=15.68\ N$

Therefore, the force of friction is 15.68 N

4 0

4 years ago

What is the the acceleration of a proton that is 4.0 cm from the center of the bead? input positive value if the acceleration is

QveST [7]

Missing detail in the text:
"<span>A small glass bead has been charged to + 25 nC "

Solution
The force exerted on a charge q by an electric field E is given by
</span> $F=qE$
<span>Considering the charge on the bead as a single point charge, the electric field generated by it is
</span> $E=k_e \frac{Q}{r^2}$
with $k_e = 8.99\cdot 10^9 Nm^2/C^2$ , $Q=+25 nC=25 \cdot 10^{-9}C$ is the charge on the bead. We want to calculate the field at $r=4.0 cm=0.04 m$ :
$E=(8.99\cdot 10^9) \frac{25\cdot 10^{-9}}{(0.04)^2}=1.4\cdot 10^5 V/m$
The proton has a charge of $q=1.6\cdot 10^{-19}C$ , therefore the force exerted on it is
$F=qE=1.6\cdot 10^{-19}C \cdot 1.4\cdot 10^5 V/m=2.25\cdot 10^{-14} N$

And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass, $m=1.67\cdot 10^{-27} kg$ , we have
$F=ma$
$a= \frac{F}{m}= \frac{2.25\cdot 10^{-14} N}{1.67\cdot 10^{-27} kg}=1.35 \cdot 10^{13} m/s^2$

The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so:
$a=-1.35 \cdot 10^{13} m/s^2$

6 0

3 years ago

Which statements about radiocarbon dating are true?

seropon [69]

Radiocarbon saying assume the half life for radioistopes change over time. The half life of C -14 is 5730 years

8 0

3 years ago

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