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3 years ago

HELP I THOUGHT IT WAS 4 AAAAAAAAAAAH

Physics

2 answers:

iren2701 [21]3 years ago

7 0

It's 2 cups per pint. I searched it up on google and that's what it said. By the way, for future reference, 4 cups is a quart and 4 quarts is a gallon.

daser333 [38]3 years ago

4 0

6 cups honey your welcome

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The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the

xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First, convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

5 0

3 years ago

Calculate the work efecttuated, when the force is 300 N, and the distance is 300 meters.

Alexxx [7]

Hello!

Use the formula:

W = Fd

Replacing:

W = 300 N * 300 m

Resolving:

W = 90000 J

The work is <u>90000 Joules.</u>

8 0

3 years ago

Read 2 more answers

Plz help!

Strike441 [17]

72km/hr = 72000m/3600seconds(we convert km/hr to m/s
= 20m/s
Velocity= 20m/s
In 4seconds,distance covered=20x4= 80m
Therefore,an accident occurs because in 4seconds, the car would have moved past 20m

6 0

3 years ago

A mass-spring system has k = 56.8 N/m and m = 0.46 kg.

andriy [413]

Answer:

A. $\omega=11.1121\ rad.s^{-1}$

B. $f=1.7685\ Hz$

C. $T=0.5654\ s$

Explanation:

Given:

spring constant, $k=56.8\ N.m^{-1}$

mass attached, $m=0.46\ kg$

for a spring-mass system the frequency is given as:

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{56.8}{0.46}}$

$\omega=11.1121\ rad.s^{-1}$

frequency is given as:

$f=\frac{\omega}{2\pi}$

$f=\frac{11.1121}{2\pi}$

$f=1.7685\ Hz$

Time period of a simple harmonic motion is given as:

$T=\frac{1}{f}$

$T=0.5654\ s$

7 0

3 years ago

Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu

11111nata11111 [884]

Answer:

velocity = 62.89 m/s in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $10^5$ kg - m/s in the right direction.