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Anna35 [415]
3 years ago
6

What can radiation do to our cells if we are exposed to it too much?

Physics
2 answers:
Debora [2.8K]3 years ago
8 0

Answer:

it can kill you cell and win you kill you no cell no life

Explanation:

rusak2 [61]3 years ago
8 0

Answer:

Our bodies are designed to deal with the low levels we're exposed to everyday due to evolution. But, too much radiation can damage tissues by changing cell structure and damaging DNA.  

Explanation:

This can cause serious health problems, including cancer.

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A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6
liraira [26]

Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

Therefore, change of speed in 5 seconds = 150 - 60

                                                                      = 90 rpm

Therefore,

90 revolutions in 1 minute

or In 1 minute the ball revolves 90 times

i.e. 1 min = 90 rev

     60 sec = 90 rev

        1 sec = 90/ 60 rec

         5 sec = $\frac{90}{60}\times 5$

                   = 75 rev

Therefore, the ball made 75 revolutions during the 5 seconds.

7 0
3 years ago
A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.
zzz [600]

Answer:

0.0257259766982 m

Explanation:

P_2 = Atmospheric pressure = 101325 Pa

d_1 = Initial diameter = 1.5 cm

d_2 = Final diameter

\rho = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m

The diameter of the bubble is 0.0257259766982 m

8 0
3 years ago
What is the lupac<br>name<br>H3C – CHT<br>the H₂ - CH - CH2<br>CH₂<br>CH₂<br>CH3​
dimulka [17.4K]

Answer:??

Explanation:

5 0
3 years ago
A ski jumper travels down a slope and
AleksandrR [38]

Answer:

304.86 metres

Explanation:

The x and y cordinates are dcos\theta and dsin\theta respectively

The horizontal distance travelled, x=v_{ox}t=dcos\theta

Making t the subject, t=\frac{dcos\theta}{v_{ox}}

Since y=0.5gt^2=dsin\theta, we substitute t with the above and obtain

0.5g(\frac{dcos\theta}{v_{ox}})^2=dsin\theta

Making d the subject we obtain

d=\frac{2v_{ox}^2sin\theta}{gcos^2\theta}

d=\frac{2*30^2sin48}{9.8cos^248}

d=304.8584

d=304.86m

5 0
3 years ago
You are testing a new amusement park roller coaster with an empty car with a mass 120 kg. One part of the track is a vertical lo
Kay [80]

Answer:

- 5436 J

Explanation:

mass of car, m = 120 kg

radius of loop, r = 12 m

velocity at the bottom (A) = Va = 25 m/s

Velocity at the top(B) = Vb = 8 m/s

Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m

by use of Work energy theorem

Work done by all the forces = change in kinetic energy of the body

Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A

- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)

- 120 x 9.8 x 24 + Work done by friction = 60 x (64 - 625)

- 28224 + Work done by friction = - 33660

Work done by friction = -33660 + 28224 = - 5436 J

4 0
3 years ago
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