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3 years ago

What are gamma rays? How are they used?

Physics

2 answers:

Vedmedyk [2.9K]3 years ago

4 0

Answer:

Gamma rays and its uses are discussed below.

Explanation:

Characteristics of Gamma rays

Radiation of short wavelength and high frequency.
Gamma rays do not deflect in a magnetic field.
Gamma rays have strong penetration energy

Uses of Gamma rays

Medical: Gamma rays are used as a medicine to kill the carcinogens cells and prevent them to grow again.
Industrial: Gamma rays are used in the industry to check the oil pipeline and discover the weak points.

Karo-lina-s [1.5K]3 years ago

3 0

I know it, X-rays, gamma rays and beta particles are all used in medicine to treat internal organs. X-rays are produced by firing electrons at a metal target and gamma rays are emitted by the nucleus of radioactive atoms. Gamma rays are used to kill cancer cells, to sterilise medical equipment and in radioactive tracers.

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At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

ozzi

Answer:

$r=2.4m$

Explanation:

We have to use the centripetal force equation

$Fc=\frac{mv^{2} }{r}$

we need the radious so we have to isolate "r" and we get

$r=\frac{mv^{2} }{Fc}$

replacing m=65 kg, v= 4.1 m/s and Fc=455N we get

$r=\frac{65*4.1^{2} }{455}$

$r=2.4m$

The radius of the amusement park chamber is 2.4m

4 0

3 years ago

An 85.0 kg fisherman jumps from a dock at a speed of 4.30 m/s onto their 135.0 kg boat. If the boat was at rest to begin but mov

jeka94

Answer:

Final speed of boat + man is 1.66 m/s

Explanation:

As we know that there is no friction on the system or there is no external force on this system

So here we can use momentum conservation here

$mv = (m + M)v_f$

so we have

m = 85 kg

M = 135 kg

v = 4.30 m/s

now we have

$85 \times 4.30 = (85 + 135) v$

$v = 1.66 m/s$

4 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$