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2 years ago

The 500 pages of a book have a mass of 2.50 kg. What is the mass of each page A in kg B in mg?

Physics

1 answer:

Furkat [3]2 years ago

3 0

Answer:

Mass per page = mass of book / number of pages

Mass per page = 2.5kg / 500 pages = 0.005kg

Mass per page = mass of book / number of pages

Total mass = 2.5kg = 2,500grams = 2,500,000mili grams

Mass per page = 2500000mg / 500 pages = 5000mg

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A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

love history [14]

Answer:

I = 16 kg*m²

Explanation:

Newton's second law for rotation

τ = I * α Formula (1)

where:

τ : It is the moment applied to the body. (Nxm)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :

ωf = ω₀+ α*t Formula (2)

Where:

α : Angular acceleration (rad/s²)

ω₀ : Initial angular speed ( rad/s)

ωf : Final angular speed ( rad

t : time interval (rad)

Data

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel

We replace data in the formula (2):

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

Magnitude of the net torque (τ )

τ = F *R

Where:

F = tangential force (N)

R = radio (m)

τ = 80 N *0.12 m

τ = 9.6 N *m

Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

I = 16 kg*m²

8 0

1 year ago

You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just

Zepler [3.9K]

Answer:

The angular magnification is $M = 2.808$

Explanation:

From the question we are told

The focal length is $f = 8.9cm$

The near point is $n_p = 25.0cm$

The angular magnification is mathematically represented as

$M = \frac{n_p}{f}$

Substituting values

$M = \frac{25}{8.9}$

$= 2.808$

4 0

1 year ago

Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo

LenaWriter [7]

Answer:

$6.33\times 10^8\ kg\cdot m/s$

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = $33\times \frac{5}{18}=9.17\ m/s$

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

= $(6.9\times 10^7\ kg)(9.17\ m/s)$

= $6.33\times 10^8\ kg\cdot m/s$

Therefore, the magnitude of ship's momentum is $6.33\times 10^8\ kg\cdot m/s$ .

6 0

2 years ago

If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal

iVinArrow [24]

Answer:

xcritical = d− m1 /m2 ( L /2−d)

Explanation: the precursor to this question will had been this

the precursor to the question can be found online.

ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)

. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces

smallest possible value of x such that the bar remains stable (call it xcritical)

∑τA = 0 = m2g(d− xcritical)− m1g( −d)

xcritical = d− m1 /m2 ( L /2−d)

6 0

1 year ago

A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with

Lana71 [14]

Answer:

h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

Em₀ = K = ½ m v₁²

final point. Highest point of the curl

$Em_{f}$ = U = m g y

Find the height y = 2R

Em₀ = Em_{f}

½ m v₁² = m g 2R

v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

-fr = m a (1)

Y Axis

N - W = 0

N = mg

the friction force has the formula

fr = μ N

fr = μ m g

we substitute 1

- μ mg = m a

a = - μ g

having the acceleration, we can use the kinematic relations

v² = v₀² - 2 a x

v₀² = v² + 2 a x

the length of this zone is x = 2R

let's calculate

v₀ = √ (4 gR + 2 μ g 2R)

v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

Em₀ = U = m g h

final point. Just before reaching the area with rubbing

$Em_{f}$ = K = ½ m v₀²

Em₀ = Em_{f}

mgh = ½ m 4gR(1 + μ)

h = ½ 4R (1+ μ)

h = 2 R (1 +μ)

7 0

2 years ago