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4 years ago

During the removal of wastes in a cell, the waste products move

Physics

1 answer:

marin [14]4 years ago

7 0

<span>From an area of higher conentration to an area of lower concentration which is B. my friend. Hoped i helped.

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A long, cylindrical solenoid with 100 turns per centimeter has a radius of 14.7 cm. If the current through the solenoid changes

oee [108]

Answer:

Explanation:

Given:

dI/dt = 6.21 A/s

n = N/l

= 100 turns/cm

= 100 turns/cm × 100 cm/1 m

Radius, r = 14.7 cm

= 0.147 m

Inductance, L = uo × n^2 × A × l

L/l = 4pi × 10^-7 × (100 × 100)^2 × pi × 0.147^2

= 8.53 H

Emf, E = L × dI/dt

E/l = L/l × dI/dt

= 8.53 × 6.21

= 52.98 V/m

4 0

3 years ago

Marissa researched the cost to have custom T-shirts printed by several local and online vendors. She found that each store’s cha

myrzilka [38]

I’m slightly confused about the wording of the question but if both the same y-intercept that means both company’s have the same flat rate charge for artwork.

8 0

3 years ago

Read 2 more answers

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

3 0

3 years ago

the impact time for a particular collision is 3.0 × 10-3 seconds and the impulse in the collision is 0.30 newton seconds. what i

Zanzabum

There are several information's already given in the question. The answer can be easily deduced using those information's.

Time = 3.0 * 10-3 seconds
Impulse = 0.30 newton
Then
Force = Impulse/Time
= 0.30/3.0 * 10-3
= 1 * 10^3 newtons.
I hope the above procedure is clear for you to understand and it has actually come to your great help.

3 0

4 years ago

Two closed organ pipe gives 4 beats when sounded together at 5°C. Calculate the number of beats in 35°C

Gre4nikov [31]

Answer:

The number of beats is 10.58 in 35°C.

Explanation:

The beat frequency is given by : f₁-f₂

At 5°C, f₁-f₂ = 4

We need to find the number of beats in 35°C.

The frequency in a standing wave is proportional to $\sqrt T$ .

So,

$\dfrac{f_1-f_2}{f_1'-f_2'}=\dfrac{\sqrt {T}}{\sqrt{T'}}\\\\f_1'-f_2'=(f_1-f_2)\times \dfrac{\sqrt{T'}}{\sqrt{T}}\\\\f_1'-f_2'=4\times \dfrac{\sqrt{35}}{\sqrt{5}}\\\\f_1'-f_2'=10.58$

So, the number of beats is 10.58 in 35°C.

7 0

3 years ago