About how many times greater is the density of a neutron star compared to a white dwarf?

A spherical bowling ball with mass m = 4.1 kg and radius R = 0.117 m is thrown down the lane with an initial speed of v = 8.9 m/

Furkat [3]

Answer:

1) 23.45 rad/s²

2) 2.7 m/s²

3) t= 1.6 s

4) x ≈ 11 m

5) vfinal = 4.45 m/s

6) KErot = 16.2 J

KEtran = 41 J

KErot < KEtran

Explanation:

Step 1: Data given

mass bowling ball = 4.1 kg

radius = 0.117 meter

initial speed = 8.9 m/s

1) What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?

α = a / r = 2.774 m/s² / 0.117m = 23.45 rad/s²

2)What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?

a = µ*g = 0.28 * 9.8m/s² = 2.744 m/s² ≈ 2.7 m/s²

3) How long does it take the bowling ball to begin rolling without slipping?

This begins when ω = v / r

with

⇒ ω = α*t = 23.45 rad/s² * t

⇒ v = Vo - a*t = 8.9m/s - 2.744m/s²*t

This gives us:

23.45rad/s² * t = (8.9m/s - 2.744m/s²*t) / 0.11m

2.744*t = 8.9 - 2.744*t

t = 8.9 / 5.488 = 1.622 s ≈ 1.6 s

4) How far does the bowling ball slide before it begins to roll without slipping?

x = Vo*t - ½at² = (8.9*1.622 - ½*2.744*(1.622)²) m = 10.82 m ≈ 11 m

5) What is the magnitude of the final velocity?

v = Vo - at = 8.9m/s - 2.744m/s² * 1.622s = 4.45 m/s

6) After the bowling ball begins to roll without slipping, compare the rotational and translational kinetic energy of the bowling ball:

trans KE = ½ * 4.1kg * (4.45m/s)² =40.595 J ≈ 41 J

I = (2/5)mr² = (2/5) * 4.1kg * (0.117m)² = 0.0224 kg·m²

ω = v/r = 4.45m/s / 0.117m = 38.03 rad/s, so

rot KE = ½Iω² = ½ * 0.0224kg·m² * (38.03rad/s)² = 16.2 J

16.2 J < 41 J

KErot < KEtran

(For a rolling solid sphere, KErot ≈ 2/5 * KEtran)

6 0

3 years ago

What factors does the speed of atoms depend on?

Vedmedyk [2.9K]

<h2>Answer:</h2>

The speed of atoms depends upon:

Temperature: Speed of atom is directly proportional to the temperature of atoms.
Pressure : It is also directly proportional to the speed of an atom.
Volume of gas : As the volume of gas increases, collision between atoms increases with increasing the speed of atom
Number of atoms surrounding: More number of atoms leads to more collisions, increasing speed of atoms.

4 0

3 years ago

A ball is dropped from a height of 10m. At the same time, another ball is thrown

soldi70 [24.7K]

5.1 m

Explanation:

Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by

$y_1 = 10 - \frac{1}{2}gt^2$ (1)

where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground $y_2,$ is given by

$y_2 = v_0t - \frac{1}{2}gt^2$ (2)

At the instant the two balls collide, they will have the same displacement, therefore

$y_1 = y_2 \Rightarrow 10 - \frac{1}{2}gt^2 = v_0t - \frac{1}{2}gt^2$

or

$v_0t = 10\:\text{m}$

Solving for t, we get

$t = \dfrac{10\:\text{m}}{v_0} = \dfrac{10\:\text{m}}{10\:\text{m/s}} = 1\:\text{s}$

We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):

$y_1 = 10\:\text{m} - \frac{1}{2}(9.8\:\text{m/s}^2)(1\:\text{s})^2$

$\:\:\:\:\:\:\:= 5.1\:\text{m}$

8 0

3 years ago

Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.

kow [346]

Answer:

a) # lap = 301.59 rad , b) L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

# turns = 1 turn (24 cm / 0.5 cm)

# laps = 48 laps

Let's reduce to radians

# laps = 48 laps (2 round / 1 round)

# lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

L₀ = 2π R

L = # turns L₀

L = # turns 2π R

L = 48 2π 30

L = 9047.79 cm

L = 90.48 m

6 0

4 years ago

An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If

wolverine [178]

Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

weight = mg

W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

taking moment about the center of the system

350*1.3=850*x

455=850x

divide both sides by 850

x=455/850

x=0.54

Hence the man must sit 0.54m from the right to balance the system

3 0

3 years ago