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3 years ago

Inflammation happens when the mucous membranes cannot stop a pathogen. True False

Physics

2 answers:

qaws [65]3 years ago

6 0

This is a true statement

mars1129 [50]3 years ago

4 0

I believe this statement is true

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Plsss helpp will give brainliest!!!

Alexandra [31]

Answer:

1. Flexibility.

2. Muscular Fitness

3. Cardiovascular Fitness

4. Cardiovascular Fitness

5. Flexibility

6. Muscular Fitness

7. Cardiovascular Fitness

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3 years ago

An electron moves with velocity v⃗ =(5.9i−6.4j)×104m/s in a magnetic field B⃗ =(−0.63i+0.65j)T. Determine the z-component of the

HACTEHA [7]

Answer:

Explanation:

Force on the electron = q ( v x B )

q = - 1.6 x 10⁻¹⁹

v = (5.9i−6.4j)×10⁴

B = (−0.63i+0.65j)

v x B = (5.9i−6.4j)×10⁴ x (−0.63i+0.65j)

= (3.835 - 4.032 ) x 10⁴ k

= - 1970 k

Force on the electron = q ( v x B )

= - 1.6 x 10⁻¹⁹ x -1970 k

= 3.152 x 10⁻¹⁶ k

z-component of the force on the electron

Fz = 3.152 x 10⁻¹⁶ N

7 0

3 years ago

Một chất điểm chuyển động trong mặt phẳng Oxy với phương trình :<br> { = 2 + 10

In-s [12.5K]

Answer:

plz write your questions in English

3 0

3 years ago

All of the alkali metals, Group 1, have one valence electron. Which of these would represent the oxidation number of the alkali

melamori03 [73]

The elements found in the Group 1, or the Alkali Metal Group, have electronic configurations that end in $s^{1}$ .This means that they have 1 electron readily available to release in order to achieve a stable state.

When these atoms release the valence electron, they will achieve a stable state. For example, Lithium's stable state will be $Li^{+}$ and Sodium will be $Na^{+}$ .

The oxidation state will then be +1.

The answer is C.

8 0

3 years ago

Read 2 more answers

The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) =

Ann [662]

Answer:

Rate of change of length as a function of time is given by $\frac{dL(t)}{dt}=34.416e^{-0.18t}$

Explanation:

The length as function of time is given by $L(t)=200(1-0.956e^{-0.18t})$

Differentiating with respect to time we get

$\frac{dL(t)}{dt}=\frac{d(200(1-0.956e^{-0.18t}))}{dt}\\\\\frac{dL(t)}{dt}=200\frac{d(1-0.956e^{-0.18t})}{dt}\\\\=200\times 0.18\times 0.956e^{-0.18t}\\\\\frac{dL(t)}{dt}=34.416e^{-0.18t}$

8 0

3 years ago