Which term is most applicable to a discussion of angular momentum in the context of black holes?

A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme

spayn [35]

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

$dy = \sqrt{d^2 - dx^2}$

$dy = \sqrt{3.85^2 - 3^2} = 2.41 km$

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

7 0

2 years ago

The relative humidity would be ________% if the actual water vapor in the air was 4 grams per cubic meter, the air's capacity to

AveGali [126]

Answer:

50%

Explanation:

Humidity is the amount water vapor present in the atmosphere.

Relative humidity is defined as the ratio of partial water vapor present in air to the actual water vapor at a particular temperature. It is expressed in percentage and the higher the percentage RH, the more the saturated water vapor present in the atmosphere and vice versa.

It is expressed mathematically as shown;

RH = actual water vapor in air/saturated water vapor × 100%

If the actual water vapor in the air was 4 grams per cubic meter and the air's capacity to hold water vapor was 8 grams per cubic meter

Actual water vapor = 4g/cm³

Air's water capacity (saturated water vapor) = 8g/cm³

RH = 4/8×100

RH = 50%

3 0

3 years ago

A truck with 28-in.-diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min, *hint convert mile

solong [7]

Answer:

Angular speed ω=3771.4 rad/min

Revolution=5921 rpm

Explanation:

Given data

$d=28in\\r=d/2=28/2=14in\\v=50mi/hr$

To find

Angular speed ω

Revolution per minute N

Solution

First we need to convert the speed of truck to inches per mile

as

1 mile=63360 inches

1 hour=60 minutes

so

$v=(50*\frac{63360}{60} )\\v=52800in/min$

Now to solve for angular speed ω by substituting the speed v and radius r in below equation

$w=\frac{v}{r}\\ w=\frac{52800in/min}{14in}\\ w=3771.4rad/min$

To solve for N(revolutions per minute) by substituting the angular speed ω in the following equation

$N=\frac{w}{2\pi }\\ N=\frac{3771.4rad/min}{2\pi }\\ N=5921RPM$

3 0

2 years ago

A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re

valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude w, pointing downward)

• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)

• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ F = -f + w sin(35°) = m a

• net perpendicular force:

∑ F = n - w cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

n = w cos(35°)

n = (15 kg) (9.8 m/s²) cos(35°)

n ≈ 120 N

Solve for the mag. of friction:

f = µ n

f = 0.25 (120 N)

f ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a

a ≈ (54.3157 N) / (15 kg)

a ≈ 3.6 m/s²

Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:

v² - v₀² = 2 a ∆x

v² = 2 (3.6 m/s²) (3 m)

v = √(21.7263 m²/s²)

v ≈ 4.7 m/s

4 0

2 years ago

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Fr

Mazyrski [523]

Answer:

x = 0.0537 m or 5.37 cm

Explanation:

Given:

spring constant'k'= 4900 N/m

radius 'r' =0.029 m

Area 'A' =r²π = 0.029²π => 2.6 x $10^{-3}$ m²

Here, Pressure 'P' is given by,

Pressure = Force / Area

And we know that, for a spring :

F = kx, where k is the spring constant and x is the change in length.

P = kx/A

As P = 101325 Pa

101325 = 4900x / ( 2.6 x $10^{-3}$ )

x = 0.0537 m or 5.37 cm

6 0

3 years ago