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3 years ago

After concluding his research, which statements would Virchow agree with? Check all that apply.

Physics

2 answers:

Viktor [21]3 years ago

8 0

Answer:

The answer is A and B

Explanation: I took the exam and i got it right

satela [25.4K]3 years ago

3 0

Answer:

(A) Living things can only come from living things.

(B) Cells come from pre-existing cells.

Explanation:

i just answered this on e d g e n u i t y.

hope this helps <3

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How does newtons 1st law apply to a project where you drop and egg?

Triss [41]

Put a parashoot on it so it touches the ground lightly and carefully

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4 years ago

A motor with a brush-and-commutator arrangement has a circular coil with radius 2.5cm and 150 turns of wire. The magnetic field

Amanda [17]

Answer:

(a) 0.81 V

(b) 0.52 V

Explanation:

Number of turns, N = 150

Radius, r = 2.5 cm = 0.025 m

Magnetic field, B = 0.060 T

f = 440 rev/min = 440 / 60 = 7.33 rps

The maximum emf is given by

e = N x B x A x 2 x π x f

e = 150 x 0.060 x 3.14 x 0.025 x 0.025 x 2 x 3.14 x 7.33

e = 0.81 V

The back emf is given by

e' = 2e / π = 2 x 0.81 / 3.14 = 0.52 V

3 0

3 years ago

What are the types of nuclear radiation ? (Particles let off by nuclear<br> materials.) *

Neko [114]

Answer:

Because nuclear decay releases ionizing radiation (alpha, beta, and gamma radiation) we call such compounds radioactive.

Explanation:

6 0

3 years ago

A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1

Afina-wow [57]

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ .

<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>

In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> ( $\vec r$ ) and <em>static</em> conditions ( $\vec r_{o}$ ) to estimate the displacement of the center of mass of the olive ( $\overrightarrow{\Delta r}$ ):

$\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right)$ (1)

Where:

$r_{i, x}$ - x-Coordinate of the i-th element of the system, in meters.
$r_{i,y}$ - y-Coordinate of the i-th element of the system, in meters.
$F_{i,x}$ - x-Component of the net force applied on the i-th element, in newtons.
$F_{i,y}$ - y-Component of the net force applied on the i-th element, in newtons.
$m_{i}$ - Mass of the i-th element, in kilograms.
$g$ - Gravitational acceleration, in meters per square second.

If we know that $\vec r_{1} = (0, 0)\,[m]$ , $\vec r_{2} = (1, 2)\,[m]$ , $\vec F_{1} = (0, 3)\,[N]$ , $\vec F_{2} = (-3, -2)\,[N]$ , $m_{1} = 0.50\,kg$ , $m_{2} = 1.50\,kg$ and $g = 9.807\,\frac{kg}{s^{2}}$ , then the displacement of the center of mass of the olive is:

<h3>Dynamic condition $\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]$ $\vec r = (0,704, 1.233)\,[m]$ </h3>

<h3>Static condition</h3><h3> $\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]$ </h3><h3> $\vec r_{o} = \left(0.75, 1.50)\,[m]$ </h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

$\overrightarrow{\Delta r} = \vec r - \vec r_{o}$

$\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]$

$\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]$

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ . $\blacksquare$

To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931

3 0

2 years ago

A girl of weight 600N sits on a 3-leg chair which weighs 60N. Each chair leg presses the floor in a circle with a diameter of 5c

Daniel [21]

Pressure = total force/total area

Total force = 660 Newton's

Total area:

Each leg contacts the floor with an area of πr^2=π(0.025m)^2=0.002m^2.

Total contact area for all 3 legs = 0.006 m^2.

Pressure = (660N) / (0.006 m^2)

= 110,000 N/m^2 = 110,000 Pascal's.

8 0

4 years ago