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A 5 kg bowling ball with a velocity of +10 m/s collides with a stationary 2 kg bowling pin. If the ball's final velocity is +8 m/s, what is the pin's final velocity?a 5 m/s
b 2.5 m/s
c 10 m/s
d 5.2 m/s

Answers

Answer 1

Answer:

Answer:

The pin's final velocity is 5m/s

Explanation:

Step one:

given data

mass of ball m1=5kg

initial velocity of ball u1=10m/s

mass of pin m2=2kg

initial velocity of pin u2= 0m/s

final velocity of ball v2=8m/s

final velocity of pin v2=?

Step two:

The expression for elastic collision is given as

m1u1+m2u2=m1v1+m2v2

substituting we have

5*10+2*0=5*8+2*v2

50+0=40+2v2

50-40=2v2

10=2v2

divide both sides by 2

v2=10/2

v2=5m/s

The pin's final velocity is 5m/s

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A spring oscillator is designed with a mass of 0.231 kg. It operates while immersed in a damping fluid, selected so that the oscillation amplitude will decrease to 1.00% of its initial value in 9.43 s. Find the required damping constant for the system.

Information that is easily converted into numbers and is stored as a on and off signals is _____ information.

Answers

"Binary" information

The index of refraction for red light in water is 1.331, and that for blue light is 1.340. If a ray of white light enters the water at an angle of incidence of 83.00o , what are the underwater angles of refraction for the blue and red components of the light

Answers

Answer:

The underwater angles of refraction for the blue and red components of the light is 47.8° and 48.2°

Explanation:

Using the Snell's law

n1 * sin Ф1 = n2 sin Ф2

1 * sin 83 = n2 sin Ф2

Ф2 = $sin^(-1) ((1)/(n2) * sin 83)$

Red light

n2 = 1.331

Ф2 = $sin^(-1) ((1)/(1.331) * sin 83) = 48.2$ °

Blue light

n2 = 1.340

Ф2 = $sin^(-1) ((1)/(1.340) * sin 83) = 47.8$ °

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade?

Answers

Incomplete question.The complete question is here

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.

Answer:

F= 0.034 N

Explanation:

Given Data

Outer=2 cm

Edge of blade=21 cm

Mass=7.7 g

Length of blade=21 cm

The last 2cm is treated as if they were at a point 20cm from the center of rotation

To Find

Force=?

Solution

Convert the given frequency to angular frequency

ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)

ω= 3/2*π rad/sec

Now to find centripetal force.

F = m×v²/r

F= m×ω²×r

Put the data

F = 0.0077 kg × (3/2×π rad/sec)²× 0.20 m

F= 0.034 N

A 2,000 kg car travels with a tangentialvelocity of 12 m/s around a circular
track with a radius of 30 meters. What
is the car's rate of centripetal
acceleration?

Answers

The car's rate of centripetal acceleration in the circular path is 4.8 m/s².

The given parameters;

mass of the car, m = 2,000 kg
velocity of the car, v = 12 m/s
radius of the circular path, r = 30 m

The centripetal acceleration of the car is calculated as follows;

$a_c = (v^2)/(r)$

where;

v is the tangential speed of the car
r is the radius of the circular path

Substitute the given parameters and solve for the centripetal acceleration;

$a_c = (12^2)/(30) \n\na_c = 4.8 \ m/s^2$

Thus, the car's rate of centripetal acceleration is 4.8 m/s².

Learn more here:brainly.com/question/11700262

a= v²/R
a = 12²/30 =4.8 m/s²

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system during the time Saturn completes half an orbit. Assume the mass of the Sun is 5.68 x10^29 kg, the mass of Saturn is 5.68 x10^26 kg, its period is 9.29 x10^8 s, and the radius of its orbit is 1.43 x 10^12 m. Ignore the influence of other celestial objects.?

Answers

Answer:

$v_(su) = 19.44 m/s$

Explanation:

$m_(su)=5.68x10^(29)kg\nm_(sa)=5.68x10^(26)kg$

$T=9.29x10^8\nr_(o)=1.43x10^(12)$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_(su)*r_(o)=(m_(sa)+m_(su))*r_(1)$

solve to r1

$r_1=(m_(sa)*r_(o))/(m_(sa)+m_(su))=(5.68x10^(26)*1.43x10^(12))/(5.68x10^(26)+5.68x10^(26))$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = (m_(sa)*2*pi*r_1^2)/(T)$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_(sun)=(m_(sa)*2*\pi *( 2r_(o)*r_1 -r_1^2))/(T)$

Since

$L_(su)= m_(su)*v_(su)*r_1$

then

$v_(su)=(m_(sa)*2*pi*(2r_(o)*r_(1)-r_(1)^2))/(T*m_(sa)*r_1)$

$v_(su) = 19.44 m/s$

Final answer:

In a two-body system such as the Sun-Saturn system, both bodies orbit around their mutual center of mass, or barycenter. Given the Sun's significantly larger mass, this barycenter is near the center of the Sun, and hence the Sun's change in velocity relative to the center of mass of the system as Saturn completes half an orbit is effectively zero.

Explanation:

The problem here is asking for the change in velocity of the Sun relative to the center of mass of the Sun-Saturn system as Saturn completes half an orbit. This is a situation involving orbital physics and center of mass systems.

However, in an isolated two-body orbit system like this, the center of mass does not change velocity - it would remain constant, not unless acted upon by an outside force, which the problem instructs us to ignore.

Saturn and the Sun both orbit around their common center of mass (their barycenter). Given that the Sun is immensely more massive than Saturn, this center of mass is located very close to the center of the Sun.

So, while the Sun does indeed move a little due to Saturn's influence, the change in velocity of Sun relative to the center of mass of the system during the time Saturn completes half an orbit, for all intents and purposes, is zero.

This is especially true unless the problem specifically mentions that the Sun is initially at rest with respect to the center of mass. In any other case, the relative velocity remains constant and hence the change is zero.

Learn more about Orbital Physics here:

brainly.com/question/12805730

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"Which gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string?"

Answers

Answer:

the second derivative of y with respect to time gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string

Explanation:

This is because the transverse wave movement of particles take place in direction 90° to direction of movement of the wave (x) itself, so second derivative of y with respect to time (t)is what will be required

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