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3 years ago

HURRY!!! I only have 14 more minutes

Physics

2 answers:

DochEvi [55]3 years ago

8 0

Answer:

B-cells

T-cells

stem cells

Im like 95% sure good luck

Explanation:

I passed biology and science

Bogdan [553]3 years ago

3 0

The correct answer is C. B-cells

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allochka39001 [22]

de mategoloalterfsqol

6 0

2 years ago

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

$V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

$PE_{f} =0.5\ PE_{i}$

5 0

3 years ago

Please help with 9 & 10

Tanzania [10]

Add all the sec. and all the meter's and then add the meter's and sec. together

4 0

3 years ago

A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.

N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

$\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2$

(c) Moment of inertia :

The net torque acting on it is, $\tau=I\alpha$ , I is the moment of inertia

Also, $\tau=Fr$

So,

$I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2$

Hence, this is the required solution.

3 0

3 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

7 0

3 years ago