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horsena [70]
3 years ago
7

People are constantly being barraged with low-level radiation. Which radiation source below contributes the most low-level radia

tion exposure to the average person?a) cosmic raysb) medical X-raysc) radond) rocks, soils, and food
Physics
2 answers:
Gennadij [26K]3 years ago
7 0

Answer:

C) Radon

Explanation:

Humans are terrified of radiation attack but we are unaware that we face radiations from a lot of natural sources which includes food and water. But their level is low and thus do not have any major impact on our body.

As per US Nuclear Regulatory Commission,on an average, an American faces a radiation of 620 millirem each year. Out of this 50% is because of Natural background radiation. Majority of this radiation is due to Radon in the air, with small contribution (around 30 millirem) from Cosmic rays and even smaller contribution from Earth itself.

Sati [7]3 years ago
3 0

I think the answer is C) Radond

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A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal
AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

v=\sqrt{v_x^2+v_y^2}           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

v_y^2=v_{oy}^2+2gh   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}

vx is calculated by using the information about the horizontal range of the ball:

R=v_o\sqrt{\frac{2h}{g}}    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0
4 years ago
In Hooke’s law, what does the x represent?
Arte-miy333 [17]
X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
7 0
4 years ago
Read 2 more answers
A 5kg box is being pulled for a force of 20 n and is sliding with an acceleration of 2 m/s. Find the coefficient of friction
RUDIKE [14]

Answer:

Coefficient of friction = 0.5

Explanation:

Given:

Mass of box = 5 kg

Force applied = 20 N

Acceleration = 2 m/s²

Find:

Coefficient of friction

Computation:

Friction force = Mass x Acceleration.

Friction force =  5 x 2

Friction force = 10 N

Coefficient of friction = Friction force / Force applied

Coefficient of friction = 10 / 20

Coefficient of friction = 0.5

3 0
3 years ago
The specific heat capacity of sea water is 4100 J/Kg°C and the boiling point of 100.6 °C. (i) Calculate the energy required to r
maw [93]

Answer:

334.314 (kJ)

Explanation:

1) the formula for the required energy is: Q=c*m(Bp-t), where c - 4100 J/kg*C; m - 0.9 kg; Bp - 100.6 C; t - 10 C.

2) according to the formula above:

Q=4100*0.9*(100.6-10)=41*9*906=334314 (J).

6 0
3 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

5 0
3 years ago
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