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2 years ago

What waves are used in the hospitals to take pictures of bones

Physics

1 answer:

stich3 [128]2 years ago

7 0

Answer:

An X-ray is used to take pictures of your bones. The waves that are used are known as radiation waves.

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Can someone please explain the quantum theory in simple terms

BaLLatris [955]

Quantum theory is the theoretical basis of modern physics that explains the nature and behavior of matter and energy on the atomic and subatomic level. The nature and behavior of matter and energy at that level is sometimes referred to as quantum physics and quantum mechanics.

6 0

2 years ago

Gravity on earth is 9.8 m/s squared, and gravity on the moon is 1.6 m/s squared. So if the mass of an object on earth is 40 kilo

garik1379 [7]

The mass of an object on Earth is the same as its mass on the Moon. The weight is different.

Weight = m * g

Weight ( Moon ) = 40 kg * 1.6 m/s² = 64 N

If the mass of an object on Earth is 40 kg, its mass on the Moon is 40 kg and its weight on the Moon is 64 N.

7 0

3 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for M, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

5 0

2 years ago

Which of these is a chemical control used to help control invasive species?

astraxan [27]

Answer:

Explanation:

Using fences.................................

4 0

3 years ago

a projectile is lunched with an initial speed of 60.0mm/s at an angle of 30.0° above the horizontal.The projectile lands on a hi

alexandr402 [8]

Answer:

52 mm/s (approximately)

Explanation:

Given:

Initial speed of the projectile is, $u=60.0\ mm/s$

Angle of projection is, $\theta=30.0\°$

Time taken to land on the hill is, $t=4\ s$

In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

$u_x=u\cos\theta\\u_x=60\times \cos(30)\\u_x=30\sqrt3\approx52\ mm/s$

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.

6 0

3 years ago