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3 years ago

Review the vocabulary associated with nuclear and wave therapies

Physics

1 answer:

tatiyna3 years ago

5 0

Answer:

Sonography- using sound waves to image internal structures

Brachytherapy- Therapy where radiatio is placed inside or next to the diseased area

lithiotripsy- The physical breakage of stones by the application of pulsed ulrasound

radiopharmaceuticals- internally delivered and targeted ionizing radiation

Explanation:

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A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The

mestny [16]

Answer:

a. Δx = 2.59 cm

Explanation:

mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m

Using momentum conserved

mb * (0) + mp * vp = ( mb + mp ) * vf

vf = ( mp / mp + mb) * vp

¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²

Solve to Δx '

Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )

Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97 m /s) ² / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]

Δx = 0.02599 m ⇒ 2.59 cm

8 0

3 years ago

19. A 0.0340 kg bullet traveling at 120 m/s embeds itself in a 1.24 kg wooden block which

Blababa [14]

Answer: x ≈ 36.3 cm

Explanation:

Conservation of momentum during the collision

0.0340(120) + 1.24(0) = (0.0340 + 1.24) v

v = 3.2025 m/s

The kinetic energy of the block/bullet mass will convert to spring potential

½kx² = ½mv²

x = √(mv²/k)

x = √(1.274(3.2025²) / 99.0)

x = 0.363293... ≈ 36.3 cm

3 0

3 years ago

Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$