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butalik [34]
3 years ago
10

Both X-rays and gamma rays have higher frequencies than ultraviolet rays True or False

Physics
2 answers:
Masja [62]3 years ago
8 0
I’m pretty sure the answer is true
mamaluj [8]3 years ago
7 0

Answer:

i believe its True

Explanation:

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Quick Physics Question
sveticcg [70]
It is Real,Virtual,The Same Size, Inverted
8 0
3 years ago
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
50 POINTS
sweet-ann [11.9K]

Answer:

The acceleration is constant and positive

Explanation:

The straight line indicates that the acceleration is constant, while the positive slope indicates that the line is positive.

7 0
3 years ago
Read 2 more answers
What do conduction, radiation, and convection work together to do?
Semmy [17]

Well, conduction is the passing of heat by touch. Convection is the passing along of heat by movement (e.g. hot air rising from a radiator to heat the rest of the room), and radiation is the heating of something through electromagnetic rays without any contact.  For example, the earth absorbs the heat radiated from the son (radiation). The air that touches the earth gets warmed up (conduction), and rises, and as it rises, it passes its heat along to more air around it (convection).

6 0
3 years ago
Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un
Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   \sigma_{f} = 570 \times \frac{3}{4}

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = \frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}

        = \frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}

       = 10.13 \times 10^{-4}

Therefore, largest size is as follows.

            Largest size = 2a

                                 = 2 \times 10.13 \times 10^{-4}

                                 = 20.26 \times 10^{-4}

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is 20.26 \times 10^{-4}.

4 0
3 years ago
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