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AlladinOne [14]

3 years ago

10

A duck is flying around and its velocity v as a function of time t is given in the graph below where rightwards is the positive

velocity direction. What time t does the duck have the same position as time t = 0 secs?

1 answer:

Gwar [14]3 years ago

6 0

Answer:

t= 2.0s

Explanation:

Khan Academy

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Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a

nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

p₀o = m₁ v₁ + m₂ v₂

pf = 0

m₁ v₁ + m₂ v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂= - (-6.2) / 4.7

m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

Xcm = 1/M (m₁ x₁ + m₂ x₂)

We derive from time

Vcm = 1/M (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

0 = 1/M (m₁ v₁ + m₂v₂)

m₁ v₁ + m₂v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂ = 1.3

4 0

3 years ago

The aorta is the main artery from the heart. a typical aorta has an inside diameter of 1.8 cm and carries blood at speeds of up

defon

Answer:

Explanation:

Volume per unit time flowing will be conserved

a₁v₁ = a₂ v₂

π r₁² x v₁ = π r₂² x v₂

(0.9 x 10⁻²)² x .35 = ( .45 x 10⁻² )² x v₂

v₂ = 1.4 m / s

3 0

3 years ago

A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in

melamori03 [73]

Answers:

a) $0.5 m/s^{2}$

b) $1.5 N$

Explanation:

a) The centripetal acceleration $a_{c}$ of an object moving in a uniform circular motion is given by the following equation:

$a_{c}=\omega^{2} r$

Where:

$\omega=1 \frac{rev}{s}$ is the angular velocity of the ball

$r=0.5 m$ is the radius of the circular motion, which is equal to the length of the string

Then:

$a_{c}=(1 \frac{rev}{s})^{2} 0.5 m$

$a_{c}=0.5 m/s^{2}$ This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force $F$ ) that is directed towards the center and is equal to the tension ( $T$ ) in the string:

$F=T=m. a_{c}$

Where $m=3 kg$ is the mass of the ball

Hence:

$T=(3 kg)(0.5 m/s^{2})$

$T=1.5 N$ This is the tension in the string

7 0

3 years ago

Find the separation of two points on the Moon’s surface that can just be resolved by the 200 in. (= 5.1 m) telescope at Mount Pa

rewona [7]

Answer:

The separation of the 2 points should be 50.0 meters.

Explanation:

According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals

$\theta =1.22\cdot \frac{\lambda }{D}$

Applying the given values we get

$\theta =1.22\cdot \frac{550\times 10^{-9} }{5.1}=0.1316\times 10^{-6}rads$

thus the linear separation equals $L=\theta \times Distance$

Applying the given values we get

$L=0.1316\times 10^{-6} \times 3.8\times 10^{5}\times 10^{3}meters\\\\\therefore L=50.00metes$

6 0

3 years ago

A woman has a mass of 55kg on earth, what is her mass on the moon

KengaRu [80]

Answer:

55 kg

Explanation:

Mass does not change and is not dependant on gravity. So even though your weight will be less on the moon your mass won't change.

7 0

3 years ago