What is the final temperature when 50 g of water at 20°c is mixed with 75 g of water at 60°c

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

The formula length x width x height is used to calculate the volume of a type of

hram777 [196]

Answer:

D

Explanation:

egwfefeewfwvebefhbdhahwuw

6 0

2 years ago

At an accident scene on a level road, investigators measure a car's skid mark to be 93 m long. It was a rainy day and the coeffi

USPshnik [31]

Answer:

u = 25 m/s

Explanation:

given,

length of skid = 93 m

coefficient of friction = 0.35

final velocity = 0 m/s

initial velocity = ?

force here is friction f = μ mg

F = ma

now com paring

-μ mg = m a

a = - μ g

a = - 0.35 x 9.8

a = -3.43 m/s²

we know,

v² = u² + 2 a s

0 = u² - 2 x 3.43 x 93

u² = 637.98

u = 25.26 m/s

u = 25 m/s (two significant figure)

6 0

3 years ago

Which of the following states that all matter tends to "warp" space in its vicinity and that objects react to this warping by ch

Natasha2012 [34]

There are no choices on the list you provided that make such a statement,
and it's difficult to understand what is meant by "the following".

That statement is one way to describe the approach to 'forces of gravity'
taken by the theory of Relativity.

4 0

3 years ago

On a bet, you try to remove water from a glass by blowing across the top of a vertical straw immersed in the water. What is the

MArishka [77]

Answer:

v₂ = 0.56 m / s

Explanation:

This exercise can be done using Bernoulli's equation

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Where points 1 and 2 are on the surface of the glass and the top of the straw

The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of the straw the velocity of the surface of the vessel is almost zero v₁ = 0

The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m

We substitute in the equation

$P_{atm}$ + ρ g y₁ = $P_{atm}$ + ½ ρ v₂² + ρ g y₂

½ v₂² = g (y₂-y₁)

v₂ = √ 2 g (y₂-y₁)

Let's calculate

v₂ = √ (2 9.8 1.6 10⁻²)

v₂ = 0.56 m / s

5 0

2 years ago