Nasa's skylab, the largest spacecraft ever to fall back to the earth, reentered the earth's atmosphere on july 11, 1979, and broke into a myriad of pieces. one of the largest fragments was a 1770-kg, lead-lined film vault, which landed with an estimated speed of 120 m/s.

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Answer 1

Answer:

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An unstable atomic nucleus of mass 1.82 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.18 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) Find the velocity of the third particle.
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A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the balloon is 0.18 kg/m3. The radius of the balloon (when filled) is R = 4.9 m. The total mass of the empty balloon and basket is mb = 121 kg and the total volume is Vb = 0.073 m3. Assume the average person that gets into the balloon has a mass mp = 73 kg and volume Vp = 0.077 m3. 1)What is the volume of helium in the balloon when fully inflated? m3 2)What is the magnitude of the force of gravity on the entire system (but with no people)? Include the mass of the balloon, basket, and helium. N
Do you think a baseball curves better at the top of a high mountain or down on a flat plain

Answer plz answer plzzz I am a little confused with full time

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I can’t read that I’m sorry make it more clear

Please helpa car is driven 200 km west and then 80 km southwest. what is the displacement of the car from the point of orgin (magnitude and direction)? draw a diagram.

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Let's take east and west to be positive and negative, respectively, and north and south to be positive and negative, respectively. Then in terms of vectors (using ijk notation), the car first moves 200 km west,

r = (-200 km) j

then 80 km southwest,

s = (-80/√2 km) i + (-80/√2 km) j

so that its total displacement is

r + s = (-80/√2 km) i + ((-200 - 80/√2) km) j

r + s ≈ (-56.6 km) i + (-256.6 km) j

This vector has magnitude

√((-56.6 km)² + (-256.6 km)²) ≈ 262.7 km

and direction θ such that

tan(θ) = (-256.6 km) / (-56.6 km) ==> θ ≈ -102.4º

relative to east, or about 12.4º west of south.

Is the magnet in a compass a permanent magnet or an electromagnet?

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the needle of a compass is a permanent magnet and the north indicator of the compass is a magnetic north pole. the north pole of a magnet lines up with the magnetic field so a suspended compass needle will rotate it lines up with the magnetic field. Answer permanent magnet

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 s after it is released. How deep is the lake

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Answer:

$|D_(depth) |=19.697m$

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

$x_(f)-x_(i)=v_(i)t+(1/2)at^(2)\n -5.0m=(o)t+(1/2)(-9.8m/s^(2) )t^(2)\n -4.9t^(2)=-5.0\n t^(2)=5/4.9\nt=√(1.02) \nt=1.01s$

As we have find the time taken now we need to find the final velocity vf from below equation as

$v_(f)=v_(i)+at\nv_(f)=0+(-9.8m/s^(2) )(1.01s) \nv_(f)=-9.898m/s$

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

$|D_(depth) |=|vt|\n|D_(depth) |=|(-9.898m/s)(1.99s)|\n|D_(depth) |=19.697m$

4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity.]

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The rate of change of angulardisplacement is defined as angular velocity. The angular velocity will be 22.41rad/s.

What is angular velocity?

The rate of change of angular displacement is defined as angular velocity. Its unit is rad/sec.

ω = θ t

Where,

θ is the angle of rotation,

tis the time

ω is the angular velocity

The given data in the problem is;

u is the initialvelocity=0

α is the angularacceleration = 4.0 rad/s²

t is the time period=

n is the number of revolution = 10 rev

From Newton's second equation of motion in terms of angular velocity;

$\rm \omega_f^2 - \omega_i^2 = 2as \n\n \rm \omega_f^2 - 0 = 2* 4 * 62.83 \n\n \rm \omega_f= 22.41 \ rad/sec$

Hence the angular velocity will be 22.41 rad/s.

To learn more about angularvelocity refer to the link

brainly.com/question/1980605

Answer:

$w_f$ = 22.41rad/s

Explanation:

First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:

$w_f^2-w_i^2=2aS$

where $w_f$ is the final angular velocity, $w_i$ the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:

$w_f^2-(0)^2=2(4)(62.83)$

Finally, solving for $w_f$ :

$w_f$ = 22.41rad/s

N5
Swinging a tennis racket against a ball is an example of a third class lever.
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Please select the best answer from the choices provided.
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Final answer:

Swinging a tennis racket against a ball as a third class lever in physics.

Explanation:

Swinging a Tennis Racket as a Third Class Lever

A tennis racket swinging against a ball is indeed an example of a third class lever in physics. In a third class lever, the effort is situated between the fulcrum and the load. In this case, the effort is provided by the player's hand gripping the racket handle, the fulcrum is the wrist joint, and the load is the ball being struck by the racket.

When a player swings the racket, the force applied by the player's hand exerts an effort on the handle of the racket. This causes the racket to rotate about the wrist joint acting as the fulcrum. The ball serves as the load, receiving the force and accelerating in the opposite direction.

Learn more about Third class lever here:

brainly.com/question/4532561

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