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ASHA 777 [7]
3 years ago
13

Please answer correctly. helppp plzzzUrgent!!!!​

Physics
1 answer:
Ksenya-84 [330]3 years ago
3 0

Answer:

Gamma rays are mostly used in the radiotherapy/ radiooncology to treat cancer. They can also be used to spot tumours. Gamma rays can kill living cells and damage malignant tumor

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Vectors a and b have scalar product â6.00, and their vector product has magnitude +9.00. what is the angle between these two vec
salantis [7]

Answer:

Value of angle between vector a and b is 56.30^{\circ}.

Explanation:

Vectors a and b have scalar product 6.00

Let \theta be the angle between a and b.

\vec{a}.\vec{b} = 6

ab cos \theta = 6 ...(1)

Vectors a and b have magnitude of vector product 9.00

\vec{a} \times\vec{b} = 9

ab sin \theta = 9 ...(2)

Dividing equation (2) by (1) we get

\frac{ab sin \theta}{ab cos \theta}  = \frac{9}{6}

tan \theta = 1.5

\theta = tan ^{-1} (1.5)

\theta = 56.30^{\circ}

Thus, value of angle between vector a and b is 56.30^{\circ}.

3 0
3 years ago
A 20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to
liberstina [14]
  • (a) Q = 1.70\times 10^{-2}\;\text{C};
  • (b) V_\text{final} = 5.31\times 10^{2}\;\text{V};
  • (c) E_\text{final} = 4.52\;\text{J};
  • (d) \Delta E = 2.82\;\text{J}.

All four values are in 3 sig. fig.

<h3>Explanation</h3>

(a)

Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}.

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial};

(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial};

\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}.

(c)

\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}.

\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}.

(d)

Initial energy of the system, which is the same as the initial energy in the 20.0\;\mu\text{F} capacitor:

\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}.

Change in energy:

\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}.

4 0
3 years ago
Cany anyone teach me how to use the vernier callipers?
luda_lava [24]
Place the object in between the two jaws such that they touch opposite ends of the object making sure the object is held firmly but don’t press too tight. If you need to measure an internal diameter, then insert the upper jaws in to the cavity and open them till they touch the sides. Tighten the locking screw to hold the jaws in position.

Note the position of the vernier scale zero on the main scale. The main scale reading is the division just before where the zero mark of the vernier scale is aligned. So is the zero mark aligns just after the fifth division between 3 and 4 the main scale reading is then 3.5.

The next step is to take the vernier scale reading. To do this find the mark on the vernier scale which lines up perfectly with a mark on the main scale. The vernier reading can then be found by multiplying the least value of the vernier scale with the number of divisions till that mark. For example if the least value is 0.01 mm and the 7thmark of the vernier scale is lined up perfectly then the vernier scale reading is 7 x 0.01 = 0.07.

The final step is to add the main scale and vernier readings to get the final measurement. For example 3.5 + 0.07 = 3.57 mm.

5 0
3 years ago
Sodium chloride, NaCl, is formed when a sodium atom transfers its electron to a chlorine atom. The difference in charge between
victus00 [196]

Sodium chloride, NaCl, is formed when a sodium atom transfers its electron to a chlorine atom. The difference in charge between the two atoms creates a(n) electrostatic attraction that bonds them together.

8 0
3 years ago
Read 2 more answers
The orbital radius of an electron in a hydrogen atom is 0.846 nm. What is its de Broglie wavelength?
kotykmax [81]

Answer:

The  value  is  \lambda   = 1.329 *10^{-9} \  m

Explanation:

From the question we are told that

  The  orbital radius is  r =  0.846nm =  0.846 *10^{-9} \ m

Generally the de Broglie wavelength is mathematically represented as

      \lambda  =  \frac{2 *  \pi  r}{4}

substituting values

     \lambda  =  \frac{ 2 * 3.142  *  0.846 *10^{-9}}{4}

    \lambda   = 1.329 *10^{-9} \  m

6 0
3 years ago
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