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4vir4ik [10]
1 year ago
13

The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV

, a
singly charged 12c+ ion moves in a circle of radius 19.4 cm in the magnetic field of a mass spectrometer. What is the magnitude of the
field? Use these atomic mass values: 12C, 12.0 u; 14C, 14.0 u; 160, 15.99 u. The conversion between atomic mass units and kilograms is
1u=1.66 x 10-27 kg.
Physics
1 answer:
Ratling [72]1 year ago
4 0

The calculated magnitude is  6.73 x 10³ V/m.

AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

Potential difference = 8.20 kV= 8.20 x 10³ V

radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

                                                                               = 1.218 m

therefore Magnitude= 8.20 x 10³ / 1.218

                                  =6.73 x 10³ V/m

Learn more about Magnitude here-

brainly.com/question/15681399

#SPJ9

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What is the force of gravity for a 12 kg turkey?<br><br> Please help asap
pogonyaev

Answer: 117.6N

Explanation:

By the second Newton's law, we know that:

F = m*a

F = force

m = mass

a = acceleration

We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.

Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:

F = 12kg*9.8m/s^2 = 117.6N

7 0
2 years ago
A 60 kg box is lifted by a rope a distance of 10 meters straight up at constant speed. how much power is required to complete th
Nadya [2.5K]
Power=Work/Time
The work done is the energy required to lift the box, fighting the force of gravity. So, Work=Potential energy of the box at 10 meters.

W=PE=mgh=(60)(9.8)(10)=5880J
Finally,
P=W/T=(5880)/(5)=1176Watt

So the answer is 1176 Watts
3 0
2 years ago
Read 2 more answers
A periodic wave with wavelength 2m has a speed of 4m/s. What is the waves frequency?.
Gekata [30.6K]

The wave frequency is 2 Hz.

What is wave frequency ?

The number of waves that pass through a fixed point in a given amount of time is referred to as the wave frequency. The hertz is the SI unit for wave frequency (Hz).

f = v / w

where,

f = frequency\\v = speed \\w = wavelength

Given,

v = 4 m/s, w = 2 m

f = 4/2 \\f = 2 Hz\\

The waves frequency is 2 Hz.

To know more about wave frequency,check out:

brainly.com/question/15830195

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5 0
1 year ago
The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. A
vagabundo [1.1K]

Answer:

Q = 1.2*10⁻¹² C

Explanation:

  • For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:

       C = \frac{Q}{V}  (1)

  • For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:

       C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d}  (2)

  • So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.
  • Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:

       Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} =  1.2e-12 C = 1.2 pC  (3)

6 0
2 years ago
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7
Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

v = \frac{I}{nqA}

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

4 0
2 years ago
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