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11 days ago

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Which statement represents a healthy choice for a pregnant woman? A mother eats twice as much to support her growing fetus. A mo

ther switches from drinking liquor to beer while pregnant. A mother with gestational diabetes decides to limit her sugar intake. A mother gains 5 to 8 pounds over the course of her pregnancy

2 answers:

Maru [1K]11 days ago

7 0

Answer:

A pregnant woman with gestational diabetes chooses to reduce her sugar consumption.

Explanation:

Maru [1K]11 days ago

3 0

Answer:

c

Explanation:

e2020

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A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at â7.0 meters per second2 to a

Keith_Richards [1034]

<span>We can deduce distance using the equation d = [(v_f^2) - (v_i^2)]/(2a). So, d = [(0^2)-(15^2)]/(2*-7) d = [0-(225)]/(-14) d = 225/14 d = 16.0714 m Considering two significant figures, the roller coaster covers approximately 16 meters as it decelerates.</span>

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11 days ago

A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction be

inna [987]

Let M denote the mass of the planet, n refer to the mass of the satellite, and r signify the radius of the planet. When the satellite is positioned at a distance r from the planet's surface, the separation between their centers is 2r. The gravitational force acting between them can be represented by the formula $f_{0} = \frac{GMm}{(2r)^{2}} = \frac{1}{4} ( \frac{GMm}{r^{2}} )$ , where G indicates the gravitational constant. If the satellite is positioned directly on the planet's surface, the distance between the two masses becomes r, and the gravitational force is represented as $f_{4} = \frac{GMm}{r^{2}} =4f_{0}$ . The answer is: $f_{4} = 4f_{0}$ .

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6 days ago

Read 2 more answers

A box weighing 46 newtons rests on an incline that makes an angle of 25° with the horizontal. What is the magnitude of the compo

inna [987]

The force of the box’s weight acting perpendicular to the slope can be computed using the formula:

F = wcos(a)

In this equation, F represents the component of the weight perpendicular to the slope,

W denotes the box's total weight,

and A is the angle of the slope.

Thus, substituting values gives: F = (46)cos(25)

Resulting in F = 42 N

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3 days ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

kicyunya [1025]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

The law of conservation of mass states that the rate of fluid mass ( $m_{1}$ ) entering a system equals the rate at which the fluid mass ( $m_{2}$ ) exits the system.

The mass flow rate can be expressed as follows:

$m = \rho A v$

where $\rho$ denotes the fluid density, $A$ signifies the cross-sectional area through which fluid flows, and $v$ represents the fluid's velocity.

Based on the problem conditions, as the fluid's density remains constant, we can write:

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas for the fluid flow, while $v_{1}$ and $v_{2}$ are the corresponding velocities across those areas.

Given the conditions in the problem, $A_{2} > A_{1}$ , leading from the formula to $v_{2} < v_{1}$ .

Furthermore, fluid pressure arises from the fluid's movement through any specific area. When the fluid accelerates, part of its energy increases its speed in the direction of flow, resulting in lower pressure.

Thus, in this instance, $v_{2} < v_{1}$ the pressure in the larger cross-sectional area $P_{2}$ will exceed the pressure $P_{1}$ in the smaller cross-sectional area, implying

$P_{2} > P_{1}$ .

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5 days ago

An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight

serg [1198]

An apple strikes the ground at a velocity of 16.2 m/s.

The angle between the velocity of the apple and a line normal to the inclined surface is 20 degrees.

The parallel and perpendicular components of its velocity concerning the surface are as follows:

$v_{perpendicular} = v cos20$

$v_{parallel} = v sin20$

This gives us:

$v_{parallel} = 16.2 sin20$

$v_{parallel} = 5.5 m/s$

The velocity along the inclined plane measures 5.5 m/s.

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10 days ago