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4 years ago

Show work and Weight - Label ALL answers! 2.2 lb = 1 kg

Physics

1 answer:

ludmilkaskok [199]4 years ago

6 0

Answer:

Bellow

Explanation:

You have the detailed explanation on the other answer

2) 6÷2.2= 2.72 kg

3) 42.5÷2.2= 19.31 kg

19.31 - 9.6= 9.71 kg difference

4) 700÷2.2 = 318.18 kg

5) now is a bit different

2.2 lbs _____1kg

X lbs _____14.6 kg

2.2×14.6 =32.12 lbs

6) 2.2×1.4=3.08 lbs

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Compare the gravitational force the sun exerts on Earth to the gravitational force Earth exerts on the sun.

podryga [215]

Answer:

Mass of sun > Mass of earth

Therefore, the sun will exert more gravitational force on earth.

Explanation:

While comparing the gravitational force exerted by two objects, we need to observe which object has a greater mass.

The object with the greater mass exerts a more gravitational force on the other object.

We know that mass of the sun is about 1.99 x 10³⁰ kg, and the earth's mass is only 6.0 x 10²⁴.

Mass of sun > Mass of earth

Therefore, the sun will exert more gravitational force on earth.

7 0

3 years ago

Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging,

Rzqust [24]

A) A. 380 kHz

To clerly see the image of the fetus, the wavelength of the ultrasound must be 1/4 of the size of the fetus, therefore

$\lambda=\frac{1}{4}(1.6 cm)=0.4 cm=0.004 m$

The frequency of a wave is given by

$f=\frac{v}{\lambda}$

where

v is the speed of the wave

$\lambda$ is the wavelength

For the ultrasound wave in this problem, we have

v = 1500 m/s is the wave speed

$\lambda=0.004 m$ is the wavelength

So, the frequency is

$f=\frac{1500 m/s}{0.004 m}=3.75\cdot 10^5 Hz=375 kHz \sim 380 kHz$

B) B. f(c+v)/c−v

The formula for the Doppler effect is:

$f'=\frac{v\pm v_r}{v\pm v_s}f$

where

f' is the apparent frequency

v is the speed of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative if it is moving away from the source)

$v_s$ is the speed of the source (positive if the source is moving away from the receiver, negative if it is moving towards the receiver)

f is the original frequency

In this problem, we have two situations:

- at first, the ultrasound waves reach the blood cells (the receiver) which are moving towards the source with speed

$v_r = +v$ (positive)

- then, the reflected waves is "emitted" by the blood cells (the source) which are moving towards the source with speed

$v_s = -v$

also

v = c = speed of sound in the blood

So the formula becomes

$f'=\frac{c + v}{v - v_s}f$

C. A. The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection

The reflection coefficient is

$R=\frac{(Z_1 -Z_2)^2}{(Z_1+Z_2)^2}$

where Z1 and Z2 are the acoustic impedances of the two mediums, and R represents the fraction of the wave that is reflected back. The acoustic impedance Z is directly proportional to the density of the medium, $\rho$ .

In order for the ultrasound to pass through the skin, Z1 and Z2 must be as close as possible: therefore, a gel with density similar to that of skin is applied, in order to make the two acoustic impedances Z1 and Z2 as close as possible, so that R becomes close to zero.

3 0

3 years ago

What do all elements in a column in the periodic table have in common?

JulsSmile [24]

Answer:

1, their atoms have the same number of valence electron. because valence electron determine the group of elements.

6 0

3 years ago

Can you please help me with this physics question

erastova [34]

Answer:

See the answers below

Explanation:

We can solve both problems using vector sum.

Let's assume the forces that help the diver dive as positive downward, and the forces that oppose upward, as negative

$F_{resultant}=100+30-85+900\\F_{resultant}=845[N]$

The drag force is horizontal d this way in the horizontal direction we will only have the drag force that produces the water stream.

$F_{drag}=50[N]$

Let's assume the forces that propel the rocket upwards as positive and forces like the weight of the rocket and other elements as negative forces.

$F_{resultant}=960+7080-7700\\F_{resultant}=340 [kN]$

6 0

3 years ago

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i

lys-0071 [83]

Answer:

W₃ = 3310.49 J , W3 = 3310.49 J

Explanation:

We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections

We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics

v2 = v₀² + 2 a₁ y

as they rest part of the rest the ricial speed is zero

v² = 2 a₁ y

a₁ = v² / 2y

a₁ = 2.3² / (2 5.90)

a₁ = 0.448 m / s²

with this acceleration we can calculate the applied force, using Newton's second law

F -W = m a₁

F = m a₁ + m g

F = m (a₁ + g)

F = 69 (0.448 + 9.8)

F = 707.1 N

Work is defined by

W₁ = F.y = F and cos tea

As the force lifts the man, this and the displacement are parallel, therefore the angle is zero

W₁ = 707.1 5.9

W₁ = 4171.89 J W3 = 3310.49 J

Let's calculate for the second part

the speed is constant, therefore they relate it to zero

F - W = 0

F = W

F = m g

F = 60 9.8

F = 588 A

the job is

W² = 588 5.9

W2 = 3469.2 J

finally the third part

in this case the initial speed is 2.3 m / s and the final speed is zero

v² = v₀² + 2 a₂ y

0 = vo2₀² + 2 a₂ y

a₂ = -v₀² / 2 y

a₂ = - 2.3²/2 5.9

a2 = - 0.448 m / s²

we calculate the force

F - W = m a₂

F = m (g + a₂)

F = 60 (9.8 - 0.448)

F = 561.1 N

we calculate the work

W3 = F and

W3 = 561.1 5.9

W3 = 3310.49 J

total work

W_total = W1 + W2 + W3

W_total = 4171.89 +3469.2 + 3310.49

w_total = 10951.58 J

4 0

3 years ago