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2 years ago

Which is more dangerous to living things: gamma rays or X-rays?

Physics

2 answers:

Dima020 [189]2 years ago

6 0

Answer:

gamma

Explanation:

x rays are often used in hospitals

gamma rays :

are highly penetrating

have highly energetic ionizing radiation

frequency is higher

wavelengths are shorter rays

can attack DNA and break the strands of that essential biological molecule

quizlet

brainly.com/question/13047731

quora

GalinKa [24]2 years ago

5 0

Grammar rays this is because they have higher penetrating effect and can only be stopped by concrete
It also has a higher ionization energy

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A center-seeking force related to acceleration is _______ force

34kurt

A because centrifugal is to velocity to how slow or fast something is and centrifugal has expresssed as ac=v2 / r (1)<span />

5 0

3 years ago

Read 2 more answers

A red laser from a physics lab is marked as producing 632.8 nm light. When light from this laser falls on two closely spaced sli

goblinko [34]

Given Information:

Wavelength of the red laser = λr = 632.8 nm

Distance between bright fringes due to red laser = yr = 5 mm

Distance between bright fringes due to laser pointer = yp = 5.14 mm

Required Information:

Wavelength of the laser pointer = λp = ?

Answer:

Wavelength of the laser pointer = λp = ?

Explanation:

The wavelength of the monochromatic light can be found using young's double slits formula,

y = Dλ/d

y/λ = D/d

Where

λ is the wavelength

y is the distance between bright fringes.

d is the double slit separation distance

D is the distance from the slits to the screen

For the red laser,

yr/λr = D/d

For the laser pointer,

yp/λp = D/d

Equating both equations yields,

yr/λr = yp/λp

Re-arrange for λp

λp = yp*λr/yr

λp = (5*632.8)/5.14

λp = 615.56 nm

Therefore, the wavelength of the small laser pointer is 615.56 nm.

3 0

3 years ago

Note that the simulation allows you to also display the force of the smaller moon

Lelu [443]

the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet

Explanation:

In this problem we are analzying the gravitational force acting between a planet and its moon.

The magnitude of the gravitational attraction between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we are considering a planet and its moon. According to Newton's third law of motion,

"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A"

If we apply this law to this situation, this means that the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.

Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

$F_{a} =\frac{F_{f}+F_{i} }{2}$ Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx : displacement

$W = F_{a} (x_{f} -x_{i} )$ :Formula (3)

$x_{f}$ : final deformation

$x_{i}$ :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

$F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N$

$F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N$

We calculate average force applying formula (2):

$F_{a} =\frac{15.4N+6.2N}{2} = 11 N$

We calculate the work done on the spring applying formula (3) : :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

$E_{pf} =\frac{1}{2} *k*x_{f}^{2}$

$E_{pi} =\frac{1}{2} *k*x_{i}^{2}$

$E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J$

$E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J$

W=ΔEp= 5.39 J-0.99 J = 4.4J

4 0

3 years ago

Help please!

Talja [164]

The correct expression for the maximum speed of the object during its motion is $\frac{FT}{m}$ .

<h3>Maximum speed of the object</h3>

The maximum speed of the object is determined using the following formulas;

v(max) = Aω

where;

A is the amplitude of the motion
ω is angular speed

The maximum speed of the object can also be obtained from the maximum net force on the object,

F = ma

where;

F is the maximum net force
a is the acceleration
m is mass of the object

F = m(v/t)

mv = Ft

v = Ft/m

Thus, the correct expression for the maximum speed of the object during its motion is $\frac{FT}{m}$ .

Learn more about maximum speed here: brainly.com/question/4931057

3 0

2 years ago