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1 year ago

Which property best makes radio waves safe for diagnosing illnesses through magnetic resonance imaging?

Physics

2 answers:

77julia77 [94]1 year ago

8 0

Answer:

1) Long energy of radio waves

2) long wavelengths of radio waves

Explanation:

aleksandrvk [35]1 year ago

4 0

Answer:

It is used in MRI because it does not damage cells

Radio waves are used for space research because they have very long wavelengths

Explanation:

Many parts of the electromagnetic spectrum are applied in clinical diagnosis and treatment of illnesses. However, these highly ionizing radiation damage cells and its dosage must be carefully managed to avoid creating radiation related health problems for the patients.

Radio waves can be used in MRI without issues because the energy of the radiation is not sufficient to cause damage to cells but is sufficient to provide images for the sake of medical diagnosis.

Secondly, radio waves have long wavelength. This property is suitable for long range

communication. Hence it can be used in space research

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In this lab, you will use a dynamics track to generate collisions between two carts. If momentum is conserved, what variable cha

BartSMP [9]

In collision type of problems since momentum is always conserved

we can say

$m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}$

So here along with this equation we also required one more equation for the restitution coefficient

$v_{2f} - v_{1f} = e(v_{1i} - v_{2i})$

so above two equations are required to find the velocity after collision

here the change in velocity occurs due to the contact force while they contact in each other

so this is the impulse of collision while they are in contact with each other while in collision which changes the velocity of two colliding objects

8 0

2 years ago

Read 2 more answers

A 2kg block of which material would require 450 joules of thermal energy to increase its temperature by 1 degree Celsius?

12345 [234]

The block is made of A) Tin, as its specific heat capacity is $0.225 J/(g^{\circ}C)$

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by $\Delta T$ , according to the following equation :

$Q=mC_s \Delta T$

where $C_s$ is the specific heat capacity of the material.

In this problem, we have:

m = 2 kg = 2000 g is the mass of the unknown material

$Q = 450 J$ is the amount of energy supplied to the block

$\Delta T = 1^{\circ}C$ is the change in temperature of the material

Solving the equation for $C_s$ , we can find the specific heat capacity of the unknown sample:

$C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)$

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

Learn more about specific heat capacity:

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7 0

1 year ago

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

$T_{mi}$ = initial temperature of metal = 100 °C

$m_{a}$ = mass of aluminum cup = 100 g

$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$

7 0

2 years ago

Calculate the first and second velocities of the car with three washers attached to the pulley, using the formulas v1 = 0.25 m /

Ber [7]

Answer:

The 1st one is 0.19

The 2nd one is 0.45

It's right for both of them :)

Explanation:

5 0

1 year ago

Read 2 more answers

Electromagnetic radiation is emitted when a charged particle moves through a medium faster than the local speed of light. This r

alexandr1967 [171]

Answer:

to create the particle the speed must be greater than 2.25 10⁸ m / s

Explanation:

In this exercise we must use the relation of the index of refraction with the speed of light in a vacuum and a material medium

n = c / v

where c is the speed of light in the vacuum, v the speed of light in the material medium and n the ratio of rafraccio

in this case they give us that the medium matter water them that has a refractive index of

n = 1,333

we clear

v = c / n

let's calculate

v = 3 10⁸ / 1,333

v = 2.25 10⁸ m / s

to create the particle the speed must be greater than 2.25 10⁸ m / s

6 0

2 years ago