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maxonik [38]
4 years ago
8

Can anything reflect x-rays or gamma rays?

Physics
2 answers:
Doss [256]4 years ago
8 0
Yes many things can reflect xrays or gamma rays including electronics ans pretty much any other matter in the world
Free_Kalibri [48]4 years ago
6 0
X-rays are reflected by Bone tissues. So, it is wrong to say that X-rays can not be reflected by anything. Gamma rays are reflected by Gamma Ray mirrors.
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A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera
jeka94

To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

h = v_0 t -\frac{1}{2} gt^2

-18 = 15*t + \frac{1}{2} 9.8*t^2

t = 3.98s

Then the total distance traveled would be

h = h_0 +v_0t

h = 18+15*3.98

h = 77.7m

Therefore the railing will be at a height of 77.7m when it has touched the ground

5 0
3 years ago
5. Draw conclusions: How are potential energy, kinetic<br> energy, and total energy related?
TEA [102]

Answer:

they are all involved in our daily lives

Explanation:

3 0
3 years ago
A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

4 0
3 years ago
4. There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 nm to 400 nm. It is not
vampirchik [111]

The ranges of frequency are:

UVA: [7.50-9.38]\cdot 10^{14} Hz

UVB: [9.38-10.71]\cdot 10^{14} Hz

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is the following:

f = \frac{c}{\lambda}

where

f is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

For the UVA, the range of wavelength is 320 nm - 400 nm, so

\lambda_1 = 320 \cdot 10^{-9} m\\\lambda_2 = 400 \cdot 10^{-9} m

So the corresponding frequencies are

f_1 = \frac{3\cdot 10^8}{320\cdot 10^{-9}}=9.38\cdot 10^{14} Hz\\f_2= \frac{3\cdot 10^8}{400\cdot 10^{-9}}=7.50\cdot 10^{14} Hz

For the UVB, the range of wavelength is 280 nm - 320 nm, so

\lambda_1 = 280 \cdot 10^{-9} m\\\lambda_2 = 320 \cdot 10^{-9} m

So the corresponding frequencies are

f_1 = \frac{3\cdot 10^8}{280\cdot 10^{-9}}=1.07\cdot 10^{15} Hz\\f_2= \frac{3\cdot 10^8}{320\cdot 10^{-9}}=9.38\cdot 10^{14} Hz

So the ranges of frequency are:

UVA: [7.50-9.38]\cdot 10^{14} Hz

UVB: [9.38-10.71]\cdot 10^{14} Hz

Learn more about waves, frequency and wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

3 0
3 years ago
If one mile is 5280 feet, how many inches are in 3.94 km?
Yakvenalex [24]

Answer:

155 118 in

Explanation:

3.94 km *  1000 m / km  *  39.37 in / m = 155 118 in

3 0
2 years ago
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