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Vaselesa [24]
3 years ago
9

Two red blood cells each have a mass of 9.0 × 10 − 14 kg and carry a negative charge spread uniformly over their surfaces. The r

epulsion arising from the excess charge prevents the cells from clumping together. One cell carries –2.50 pC of charge and the other –3.10 pC, and each cell can be modeled as a sphere 7.5 μm in diameter. (a) What speed would they need when very far away from each other to get close enough to just touch? Ignore viscous drag from the surrounding liquid. (b) What is the magnitude of the maximum acceleration of each cell in part (a)?
Physics
1 answer:
Gennadij [26K]3 years ago
8 0

Answer:

321.39 m/s

1.3772213803 \times 10^{10}\ m/s^2

Explanation:

m = Mass = 9\times 10^{-14}\ kg

q_1 = Charge = -2.50 pC

q_2 = Charge = -3.10 pC

r = Radius = 3.75 μm

\varepsilon_{0} = Vacuum permittivity = 8.85 \times 10^{-12}\ F/m

We have the relation

m v^{2} &=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} 2 r}

v =\sqrt{\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}(2 r)(m)}}

=\sqrt{\frac{\left(-2.50 \times 10^{-12}\right)\left(-3.1 \times 10^{-12}\right)}{4(3.14)\left(8.85 \times 10^{-12}\right)\left(2 \times 3.75 \times 10^{-6}\right)\left(9 \times 10^{-14}\right)}}

=321.39\ m/s

The Speed they need when very far away from each other to get close enough to just touch is 321.39 m/s

We have the relation

m a=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}(2 r)^{2}}

a=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}(2 r)^{2}(m)}

=\frac{\left(-2.50 \times 10^{-12}\right)\left(-3.10 \times 10^{-12}\right)}{4(3.14)\left(8.85 \times 10^{-12}\right)\left(2 \times 3.75 \times 10^{-6}\right)^{2}\left(9 \times 10^{-14}\right)}

a=1.3772213803 \times 10^{10} \mathrm{m} / \mathrm{s}^{2}

The acceleration is 1.3772213803 \times 10^{10}\ m/s^2

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A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is. (a) How m
Gelneren [198K]

Answer:

a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C

Explanation:

Here is the complete question

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?

Solution

a.

i = Q/t = ne/t

n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s

So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C

       = 4.98 × 10¹⁹ protons

       ≅ 5 × 10¹⁹ protons

b

The total kinetic energy of the protons = heat change of target

total kinetic energy of the protons = n × kinetic energy per proton

                                                         = 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton

                                                         = 30 × 10⁷ J

heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)

ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)

     = 30 × 10⁷/14.62

     = 2.05 × 10⁷ °C

5 0
3 years ago
An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

5 0
3 years ago
Help me please I can't get the final step​
inna [77]

Answer:

\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

\displaystyle A=\frac{3}{2}B^mC^n

and the dimensions of each variable is:

A=L^2T^2

B=LT^{-1}

C=LT^2

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n

Operating:

L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)

L^2T^2=L^{m+m}T^{-m+2n}

Equating the exponents:

m+n=2

-m+2n=2

Adding both equations:

3n=4

Solving:

n=4/3

m=2-4/3=2/3

Answer:

\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}

6 0
3 years ago
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