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3 years ago

The real advantage to hydrostatic weighing is that it __________.

Physics

2 answers:

katovenus [111]3 years ago

7 0

The real advantage to hydrostatic weighing is that it gives one of the most accurate measurements of body fat.

KengaRu [80]3 years ago

7 0

Answer:

yup its B as in bagel

Explanation:

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Statement A: Area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

pochemuha

Explanation:

Formula for calculating the area of a rectangle A = Length *width

For statement A;

Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

Area of the rectangle = 2.536mm * 1.4mm

Area of the rectangle = 3.5504mm²

The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.

Area of the rectangle = 3.6mm² (to 2sf)

For statement B;

Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.

Area of the rectangle = 2.536mm * 1.41mm

Area of the rectangle = 3.57576mm²

Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.

Area of the rectangle = 3.58mm² (to 3sf)

Based on the conversion, it can be seen that 3.6mm² is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.

7 0

3 years ago

Two toy cars (m1 = 0.200 kg and m2 = 0.250 kg) are held together rear to rear with a compressed spring between them. When they a

timurjin [86]

Answer:

Acceleration of the second particle at that moment is given as

$a_2 = 2.12 m/s^2$

Explanation:

As we know that both cars are connected by same spring

So on this system of two cars there is no external force

So we will have

$F = 0 = m_1a_1 + m_2a_2$

now we have

$m_1 = 0.200 kg$

$m_2 = 0.250 kg$

$a_1 = 2.65 m/s^2$

now we have

$0.200(2.65) + 0.250a_2 = 0$

so we have

$a_2 = 2.12 m/s^2$

8 0

3 years ago

Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs

Temka [501]

The motion of the ball on the vertical axis is an accelerated motion, with acceleration
$a=g=-9.81 m/s^2$
The following relationship holds for an uniformly accelerated motion:
$2aS=v_f^2 - v_i^2$
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
$v_f =0$
So we can rewrite the equation as
$2(-9.81 m/s^2) h=-v_i^2$
from which we can isolate h
$h= \frac{v_i^2}{19.62}$ (1)

Now let's assume that $v_i$ is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: $2v_i$ . So the maximum height of the second ball is
$h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62}$ (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.

8 0

3 years ago

With each bounce off the floor, a tennis ball loses 11% of its mechanical energy due to friction. When the ball is released from

Reika [66]

Answer:

240 cm

Explanation:

Gradpoint

3 0

3 years ago

Help please I'll mark as brainliest for the first correct answer

Alenkinab [10]

Answer:

first one i think is this. work = 1/2 kx^2 = 1/2 Fx

2nd, is 0.08 J

Explanation:

EE = ½ kx²

EE = ½ (400 N/m) (0.02 m) ²

EE = 0.08.

THIRD, Velocity of the stone is 4 m/s when it leaves catapult.

3 0

3 years ago