Plates move mainly due to

Is water actually wet?

pychu [463]

yes

Explanation:

water contains water to prove it is wet

5 0

3 years ago

Read 2 more answers

A plank rests on top of the axles of two identical wheels. Each wheel's outer radius is 0.25 meters and each wheel's axle has ra

kaheart [24]

Answer:

The plank moves 0.2m from it's original position

Explanation:

we can do this question from the constraints that ,

the wheel and the axle have the same angular speed or velocity
the speed of the plank is equal to the speed of the axle at the topmost point .

thus ,

since the wheel is pure rolling or not slipping,

⇒ $v=wr$

where

$v$ - speed of the wheel

$w$ - angular speed of the wheel

$r$ - radius of the wheel

since the wheel traverses 1 m let's say in time ' $t$ ' ,

$v_{w}=\frac{distance}{time} =\frac{1}{t}$

∴

⇒ $w=\frac{v_{w}}{r} = \frac{1}{t*0.25}$

the speed at the topmost point of the axle is :

⇒ $v_{a}=w*r\\v_{a}=\frac{1}{t*0.25} *0.05\\v_{a}=\frac{1}{5t}$

this is the speed of the plank too.

thus the distance covered by plank in time ' $t$ ' is ,

⇒ $d=v_{a}*t\\d=\frac{1}{5t} *t\\d=\frac{1}{5} = 0.2m$

8 0

4 years ago

An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m

Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

40100 = ½ (47.5 + 34.3)t

t = 980 s

7 0

4 years ago

When the effect of aerodynamic drag is included, the y‐acceleration of a baseball moving vertically upward is au = −g − kv2, whi

alexira [117]

Answer:

Explanation:

The detailed steps and appropriate integration and differentiation is as shown in the attached files.

au = −g − kv2

ad = −g + kv2

3 0

3 years ago

The Earth orbits the Sun with a period of 3.16 x 107 seconds (1 year!) at a distance of 1.5 x 1011 m. The mass of the Earth is 6

SVETLANKA909090 [29]

Answer:

a) M = 2 10³⁰ kg , b) $T_{m}$ = 5.81 10⁷ s

Explanation:

a) For this exercise let's use Newton's second law where force is the law of universal gravitation and acceleration is centripetal

G m M / R² = m a

a = v² / R

G M / R = V²

The orbit of the two planets is approximately circular, therefore the velocity module (speed) is constant

v = d / t

The distance is the length of the circular orbit

d = 2π R

G M / R = 4π² R² / T²

G M T² = 4π² R³

Let's write this equation for each planet

For the earth

The period is T = 3.16 10⁷ s and the radius of the orbit R = 1.5 10¹¹ m, let's calculate the mass of the sun

M = 4π² R³ / G T²

M = 4π² (1.5 10¹¹)³ / (6.67 10⁻¹¹ (3.16 10⁷)²)

M = 133.24 10³³ / 66.60 10³

M = 2 10³⁰ kg

b) For this part we write this equation for the two points

For the earth

$T_{E}$ ² = (4π² / G M) $R_{E}$ ³

For mars

$T_{m}$ ² = (4π² / G M) $R_{m}$ ³

Let's divide the two expressions

$T_{m}$ ² / $T_{E}$ ² = $R_{m}$ ³ / $T_{E}$ ³

They indicate that the orbit of Mars is $R_{m}$ = 1.5 $R_{E}$

$T_{m}$ ² / $T_{E}$ ² = (1.5 $R_{E}$ / $R_{E}$ )³

$T_{m}$ ² = $T_{E}$ ² 1.5³

$T_{m}$ ² = (3.16 10⁷)² 1.5³

$T_{m}$ = √ (33.70 10¹⁴)

$T_{m}$ = 5.81 10⁷ s

5 0

3 years ago