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3 years ago

11

How popsicles are made?

1 answer:

crimeas [40]3 years ago

6 0

It is made<span> by freezing flavored liquid (such as fruit juice) around a stick, generally resembling a tongue depressor. Often, the juice is colored artificially. Once the liquid freezes solid, the stick can be used as a handle to hold the ice pop</span>

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A cylindrical blood vessel is partially blocked by the buildup of plaque. At one point, the plaque decreases the diameter of the

densk [106]

Answer:10.4 times of initial velocity

Explanation:

Given

Diameter reduced by 69 %

it approaches with velocity $v_0$

suppose its velocity is v during blocked passage

suppose d is the initial diameter and $d_2$ diameter is

$d_2=d(1-0.69)$

$d_2=0.31 d$

$A_2=\frac{\pi d_2^2}{4}$

As flow is constant

$Q_1=Q_2$

$d^2v_0=d_2^2v$

$v=10.40 v_0$

6 0

3 years ago

The data in the table above show how far

Taya2010 [7]

Answer:c

Explanation: because if you work it in a paper it should like lil wit is straight the numbers are going up by 16

6 0

3 years ago

How can voltage be induced in a wire with the help of a magnet?

babunello [35]

If you move a magnet through a loop of wire, induction will happen. The more loops you make, the stronger the effect becomes.

3 0

3 years ago

Carter found the rock shown in the picture below near a river

noname [10]

Answer:

there's no picture

Explanation:

and what would the question be anyways?

4 0

3 years ago

Read 2 more answers

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

$J=\frac{E}{\rho}$ (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

$J=\frac{I}{A}$ (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

$\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}$

Next, you replace for all variables:

$I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A$

hence, the current in the wire is 4.75A

4 0

3 years ago