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Alexxandr [17]
3 years ago
7

What is the cancer treatments

Physics
2 answers:
Misha Larkins [42]3 years ago
5 0
Chemo therapy which uses radiation
ad-work [718]3 years ago
3 0

One is chemo. Chemo is a special magnetic field like to treat cancer


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how does distance change when the amount of effort force is increasesd while the amount of work done remains the same?
nlexa [21]
Could you supply u with some answer choices please?
8 0
4 years ago
How much current would flow in a device powered by four 1.5 V batteries through a bulb that had a resistance of 51.0 ohms?
Serhud [2]

Answer:

0.12A

Explanation:

Total voltage produced by the batteries;

V = 1.5 × 4 = 6 v

Then the current that is flowing through the device

Current(I) = V/R

; I = 6/51

;Current = 0.12A

4 0
3 years ago
If the mass of a material is 122 grams and the volume of the material is 10 cm3, what would the density of the material be?
sergij07 [2.7K]
<span>Since d=m/v. 122 g (mass) divided by 10 cm3 (volume)=</span>12.2 g/cm<span>3
Your answer would be 12.2 g/cm3
</span>
3 0
4 years ago
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo
Vitek1552 [10]

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x d}{L}

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = \frac{4.53\ cm}{10} = 0.00453 m

Therefore,

\lambda = \frac{(0.00453\ m)(0.00109\ m)}{8.61\ m}

<u>λ = 5.734 x 10⁻⁷ m = 573.4 nm</u>

6 0
3 years ago
A 2.00 kg, frictionless block s attached to an ideal spring with force constant 550 N/m. At t = 0 the spring is neither stretche
gladu [14]

Answer:

a)    A = 0.603 m , b) a = 165.8 m / s² , c)  F = 331.7 N

Explanation:

For this exercise we use the law of conservation of energy

Starting point before touching the spring

    Em₀ = K = ½ m v²

End Point with fully compressed spring

    Em_{f} = K_{e} = ½ k x²

    Emo = Em_{f}

    ½ m v² = ½ k x²

    x = √(m / k)    v

    x = √ (2.00 / 550)   10.0

    x = 0.603 m

This is the maximum compression corresponding to the range of motion

     A = 0.603 m

b) Let's write Newton's second law at the point of maximum compression

    F = m a

    k x = ma

    a = k / m x

    a = 550 / 2.00 0.603

    a = 165.8 m / s²

With direction to the right (positive)

c) The value of the elastic force, let's calculate

    F = k x

    F = 550 0.603

   F = 331.65 N

5 0
3 years ago
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