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Licemer1 [7]
3 years ago
8

Which particle is used as a beam to treat cancer? electron neutron proton

Physics
2 answers:
melamori03 [73]3 years ago
6 0

Answer:

Proton

Explanation:

Proton therapy, also called proton beam therapy, is a type of radiation therapy. It uses protons rather than x-rays to treat cancer. A proton is a positively charged particle. At high energy, protons can destroy cancer cells.

hammer [34]3 years ago
4 0

Answer:neutron

Explanation:

Because it is similar to the motor neurons

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Fill in the correct answers. A big wheel with twice the circumference of a small wheel will rotate with ____ the force and ___ t
Lina20 [59]

Answer:

C.

Explanation:

""""""" Twice,Half """"""

3 0
3 years ago
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Water ______________ easier in the mountains because there is less air pressure.
m_a_m_a [10]

Answer:

water flows easier in the mountains because there is less air pressure.

6 0
2 years ago
A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ
DochEvi [55]

Answer:

a)  k=19.6N/m

b)  V_m=0.81m/s

c)  a_m=6.561m/s^2

d)  K.E=0.096J

e)  T=0.78sec &F=1.29sec

f)   mx'' + kx' =0

Explanation:

From the question we are told that:

Stretch Length L=0.150m

Mass m=0.30kg

Total stretch lengthL_t=0.150+0.100=>0.25

a)

Generally the equation for Force F on the spring is mathematically given by

F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}

k=19.6N/m

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

V_m=A\omega

Where

A=Amplitude

A=0.100m

And

\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s

Therefore

V_m=A\omega\\\\V_m=8.1*0.1

V_m=0.81m/s

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

a_m=\omega^2A

a_m=8.1^2*0.1

a_m=6.561m/s^2

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*0.3*0.8^2

K.E=0.096J

e)

Generally the equation for  the period T is mathematically given by

\omega=\frac{2\pi}{T}

T=\frac{2*3.142}{8.1}

T=0.78sec

Generally the equation for  the Frequency is mathematically given by

F=\frac{1}{T}

F=1.29sec

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

mx'' + kx' =0

Where

'= signify order of differentiation

7 0
3 years ago
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed
sergij07 [2.7K]

Answer:

v_{B0}=15.73 m/s

Explanation:

We can use the conservation of momentum. The initial momentum is equal to the final momentum:

x-coordinate

p_{0x}=p_{fx}

m_{A}v_{A0}=(m_{A}+m_{B})v_{cx}  

m_{A}v_{A0}=(m_{A}+m_{B})v_{c}cos(40) (1)

y-coordinate

p_{0y}=p_{fy}

m_{B}v_{B0}=(m_{A}+m_{B})v_{cy}  

m_{B}v_{B0}=(m_{A}+m_{B})v_{c}sin(40) (2)

We can divide equations (2) and (1):

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=\frac{sin(40)}{cos(40)}

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=tan(40)

v_{B0}=\frac{m_{A}v_{A0}}{m_{B}}*tan(40)

v_{B0}=\frac{1200*25}{1600}*tan(40)

v_{B0}=15.73 m/s

I hope it helps you!

           

4 0
3 years ago
Read 2 more answers
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