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3 years ago

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A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 0.200 fg (where a femtogra

m, fg, is 10^−15g) and is swimming at a velocity of v = 1.00 μm/s , with an uncertainty in the velocity of 5.00 % . E. coli bacterial cells are around 1 μm (10^−6 m) in length. The student is supposed to observe the bacterium and make a drawing. However, the student, having just learned about the Heisenberg uncertainty principle, complains that she cannot make the drawing. She claims that the uncertainty of the bacterium's position is greater than the microscope's viewing field, and the bacterium is thus impossible to locate.
1. What is the uncertainty of the position of the bacterium?

1 answer:

andriy [413]3 years ago

3 0

Answer:

58.22pm

Explanation:

Look up attached file

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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th

Darya [45]

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = $\frac{v^2}{r}$ [v = linear velocity, r = radius of circular path]

=> F = m $\frac{v^2}{r}$ ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m $\frac{v^2}{r}$ = q v B sin θ [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m $\frac{v^2}{r}$ = q v B sin 90°

=> m $\frac{v^2}{r}$ = q v B

Divide both side by v;

=> m $\frac{v}{r}$ = qB

Make v subject of the formula

v = $\frac{qBr}{m}$

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = $\frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}$

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = $\frac{1}{2}$ mv²

m = mass of proton

v = velocity of the proton as calculated above

K = $\frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )$

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

8 0

3 years ago

You are an astronaut in space far away from any gravitational field, and you throw a rock as hard as you can. The rock will:

Nesterboy [21]

Answer:

the rock will continue at the same speed unless it is affected by another force such as gravity and so if you threw it it will continue to move unless affected by a force

Explanation:

this is because Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

7 0

3 years ago

A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr

Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

$F_{x}=-1N.m$

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

$T=r*F$

Where

$F=F_{x}i+(7N)j-(5N)k$ and the position vector

$r=(2m)i-(3m)j+(2m)k$

using the determinant method to expand the cross product in order to determine the torque we have

$\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\$

by expanding we arrive at

$T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\$

since we have determine the vector value of the toque, we now compare with the torque value given in the question

$(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\$

if we directly compare the j coordinate we have

$10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m$

8 0

3 years ago

A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the

valina [46]

The speed of water can be split into vertical and horizontal speed components:
$v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta$

Due to the force of gravity, the y component will be parabolic. The x component will be linear:
$y(t) = -4.9t^2 + (6.5sin \theta) t \\ \\ x(t) = (6.5 cos \theta) t$
To find when the water hits the ground 2.5m away, set y= 0 and x = 2.5
$-4.9t^2 + (6.5sin \theta) t=0 \\ \\ t = \frac{6.5}{4.9} sin \theta \\ \\(6.5 cos \theta)(\frac{6.5}{4.9} sin \theta) = 2.5 \\ \\ sin \theta cos \theta = 0.29 \\ \\ sin 2\theta = 0.58 \\ \\ 2\theta = 35.4, 144.6 \\ \\ \theta = 17.7,72.3$

8 0

3 years ago

A mass m is attached to a string connected to a force sensor on a rotating platform. The platform’s angular velocity, ω, can be

makkiz [27]

Answer:

The mass m is 0.332 kg or 332 gm

Explanation:

Given

The platform is rotating with angular speed , $\omega =8.5\, \frac{rad}{sec}$

Mass m is moving on platform in a circle with radius , $r=0.20\, m$

Force sensor reading to which spring is attached , $F=4.8\, N$

Now for the mass m to move in circle the required centripetal force is given by $F=m\omega ^{2}r$

=> $4.8=m\times 8.5 ^{2}\times 0.20$

$=>m=0.332\, kg$

Thus the mass m is 0.332 kg or 332 gm

7 0

3 years ago