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3 years ago

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Why does an object in motion stay in motion unless acted on by an unbalanced force? A) because forces are what stop and start mo

tion B) because balanced forces can not exist in nature C) because motion is the inherent state of all matter Eliminate D) because motion, once begun, continues to accelerate

1 answer:

ruslelena [56]3 years ago

3 0

The answer is A you are welcome

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How do crumple zones extend the deceleration of a car?

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Answer:

Uv rays

Explanation:

ultraviolet rays of the suun gives force

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3 years ago

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A typical adult human lung contains about 330 million tiny cavities called alveoli. Estimate the average diameter of a single al

Amanda [17]

Answer:

The average diameter of a single alveolus is 0.0222 cm.

Explanation:

Volume of the lung ,V= 1.9 L

$1 L = 1000 cm^3$

$1.9 L=1.9\times 1000 cm^3=1900 cm^3$

Number of alveoli in a human lung = $330\times 10^6$

Volume of single alveoli =v

$v\times 330\times 10^6=V$

$v=\frac{1900 cm^3}{330\times 10^6}$

$v=5.7575\times 10^{-6} cm^3$

The alveoli are spherical.

Radius of an alveolus = r

Volume of the sphere = $\frac{4}{3}\pi r^3$

$v=\frac{4}{3}\pi r^3$

$5.7575\times 10^{-6} cm^3=\frac{4}{3}\times 3.14\times r^3$

$r=0.0111 cm$

Diameter of the alveolus =d

d = 2r = 2 × 0.0111 cm = 0.0222 cm

The average diameter of a single alveolus is 0.0222 cm.

7 0

3 years ago

A flashlight can be made that is powered by the induced current from a magnet moving through a coil of wire. The coil and magnet

rodikova [14]

Answer:

n = 4.25 x 10⁴ .

Explanation:

EMF required to be produced = 3 V .

Let n be the required no of turns .

rate of change of flux = n x B x A x f where n is no of turns of coil , B is magnetic field , A is area of coil and f is frequency of change .

B = .05 T , A = π x (.015)² = 7.065 x 10⁻⁴ m² , f = 2

emf induced = rate of change of flux

= n x .05 T x 7.065 x 10⁻⁴ m² x 2

= .7065 x 10⁴ x n

Given

.7065 x 10⁻⁴ x n = 3

n = 4.25 x 10⁴ .

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3 years ago

What term is used to describe DNA Replication?

ad-work [718]

Answer:

Semi-conservative

Explanation:

Each strand of the original DNA molecule serves as a template for the production of its counterpart

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3 years ago

A charged nonconducting rod, with a length of 3.06 m and a cross-sectional area of 4.32 cm2, lies along the positive side of an

Radda [10]

Answer:

(a): $\rm 3.560\times 10^{10}\ electrons$

(b): $\rm 6.675\times 10^{10}\ electrons$

Explanation:

<u>Given:</u>

Length of the rod, L = 3.06 m.
Cross-sectional area of the rod, A = 4.32 cm² = $\rm 4.32\times 10^{-4}\ m^2.$ .

The charge on a single electron, $\rm e=-1.9\times 10^{-19}\ C.$

The excess number of electrons on the rod is equal to the net charge on the rod divided by the charge on a single electron.

(a): If the volume charge density of the rod is uniform.

Uniform volume charge density of the rod, $\rm \rho = -4.31\ muC/m^3= -4.31\times 10^{-6}\ C/m^3.$

For the uniform charge density, the total charge on the rod is equal to the product of volume of the rod and the volume charge density of the rod.

$\rm Q=\rho \times Volume\ of\ the\ rod\\=\rho\times (A\times L)\\=-4.31\times 10^{-6}\times (4.32\times 10^{-4}\times 3.06)\\=-5.697\times 10^{-9}\ C.$

If there are n number of excess electron on the rod, then,

$\rm Q=ne\\n=\dfrac Qe =\dfrac{-5.697\times 10^{-9}}{-1.6\times 10^{-19}}=3.56\times 10^{10}\ electrons.$

(b): If the volume charge density of the rod is non-uniform.

Non-uniform volume charge density of the rod, $\rm \rho(x) = bx^2.$

where, $\rm b=-2.59\ muC/m^5= -2.59\times 10^{-6}\ C/m^5.$

For the non-uniform charge density of the rod, the total charge on the rod is given by

$\rm Q=\int \rho(x)\ dV.$

where, dV is the volume element of the rod whose length is dx, therefore, dV = A dx.

The rod is extended from origin to length L, therefore, the total charge on the rod is given by,

$\rm Q = \int\limits^{L}_{0} \rho(x)A\ dx\\= \int\limits^{L}_{0} bx^2A\ dx\\= bA\int\limits^{L}_{0} x^2\ dx\\=bA\left (\dfrac{x^3}3 \right )\limits^{L}_{0}\\=bA\dfrac{L^3}{3}\\=\dfrac 13 \times (-2.59\times 10^{-6})\times(4.32\times 10^{-4})\times(3.06^3) \\=-1.068\times 10^{-8}\ C.$

If there are n number of excess electron on the rod, then,

$\rm Q=ne\\n=\dfrac Qe =\dfrac{-1.068\times 10^{-8}}{-1.6\times 10^{-19}}=6.675\times 10^{10}\ electrons.$

8 0

3 years ago