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elena55 [62]
3 years ago
11

On what criteria is the Fujita scale for Tornadoes based?

Physics
1 answer:
svp [43]3 years ago
7 0
B. Tornado destruction
It is based on the amount of damage
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6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.
Sidana [21]

Answer:

acceleration of the rocket is given as

a = 7.45 m/s^2

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

v_i = 0

v_f = 447 m/s

t = 1 min = 60 s

so we have

a = \frac{v_f - v_i}{t}

a = \frac{447 - 0}{60}

a = 7.45 m/s^2

7 0
2 years ago
If b humans cannot see ultraviolet waves how can ultraviolet light be used to gather evidence of a crime?
Inga [223]
I don’t know because it’s very difficult
6 0
3 years ago
If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density?
Alex Ar [27]

Answer:

They have same density

Explanation:

The density of an object is defined as

d=\frac{m}{V}

where

m is the mass of the object

V is its volume

Let's call m_c and V_c the mass and the volume of ball C, respectively. Therefore, the density of ball C is:

d_c = \frac{m_c}{V_c}

We know that the volume of ball C is 3 times the volume of ball D, so

V_c = 3 V_d \rightarrow V_d = \frac{V_c}{3}

And we also know that ball D has 1/3 the mass of ball C:

m_d = \frac{m_c}{3}

So, the density of ball D is:

v_d = \frac{m_d}{V_d}=\frac{m_c/3}{V_d/3}=\frac{m_c}{V_c}=d_c

Therefore, the two balls have same density.

6 0
3 years ago
Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th
gregori [183]

Complete Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is 1.28 * 10^9years.

Answer:

The potassium-40 present in 80 kg is  Z = 0.0288 *10^{-3}\ kg

The effective dose absorbed per year is  x = 2.06 *10^{-24} per year

Explanation:

From the question we are told that

      The mass of potassium in 1 kg of human body is m =  3g= \frac{3}{1000} =  3*10^{-3} \ kg

      The mass of the person is M = 80 \ kg

       The abundance of Potassium-39 is   93.26%

        The abundance of Potassium-40 is   0.012%

         The abundance of Potassium-41 is   6.78 %

         The energy absorbed is  E =  1.10MeV = 1.10 *10^{6} * 1.602 *10^{-19} = 1.7622*10^{-13} J

Now  1 kg of human body contains       3.0*10^{-3}\ kg of  Potassium

So      80 kg of human body contains      k kg of  Potassium

=>   k = \frac{ 80 * 3*10^{-3}}{1}

     k = 0.240\  kg

Now from the question potassium-40 is  0.012% of the total  potassium so

     Amount of potassium-40  present is mathematically represented as

            Z = \frac{0.012}{100}  * 0.240

            Z = 0.0288 *10^{-3}\ kg

The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as

           D =  \frac{E}{M}

Substituting values

          D =  \frac{1.7622*10^{-13}}{80}

            D =  2.2*10^{-15} J/kg

Converting to Sieverts

We have

           D_s = REB * D

           D_s = 1.2 * 2.2 *10^{-15}

           D_s =  2.64 *10^{-15}

So

     for half-life (1.28 *10^9 \ years)  the dose is  2.64 *10^{-15}

     Then for 1  year the dose would be  x

=>         x = \frac{2.64 *10^{-15}}{1.28 * 10^9}

             x = 2.06 *10^{-24} per year      

7 0
3 years ago
A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the
Umnica [9.8K]

Answer:

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

H_{max}=\frac{V^2\sin ^2 \theta}{2g}

g = 9.8m/s2

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

6 0
3 years ago
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