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3 years ago

What is the best description of a mechanical wave?

Physics

1 answer:

gizmo_the_mogwai [7]3 years ago

8 0

Answer:

A mechanical wave is a wave that is an oscillation of matter, and therefore transfers energy through a medium. While waves can move over long distances, the movement of the medium of transmission—the material—is limited. Therefore, the oscillating material does not move far from its initial equilibrium position.

Explanation:

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A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th

ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

5 0

3 years ago

An object that is initially not rotating has a constant torque of 3.6 N⋅m applied to it. The object has a moment of inertia of 6

Korolek [52]

Answer:

0.6

Explanation:

Angular acceleration is equal to Net Torque divided by rotational inertia, which is the rotational equivalent to Newton’s 2nd Law. Therefore, angular acceleration is equal to 3.6/6 which is 0.6. Hope this helped!

3 0

2 years ago

A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

$A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}$

Differentiate both sides with respect to d.

$\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)$

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

$\frac{d^{2}A}{dt^{2}}= + ve$

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0

3 years ago

It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T

tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

$Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\$

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

$\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\$

7 0

3 years ago

A red train travelling at 72 km/h and a green train travelling at 144 km/h are headed toward each

abruzzese [7]

Answer:

Collision will occur.

Speed of red train when they collide = 0 m/s.

Speed of green train when they collide = 10 m/s.

Explanation:

Speed of red train = 72 km/h = 20 m/s

Speed of green train = 144 km/h = 40 m/s.

Deceleration of trains = 1 m/s²

For red train:-

Equation of motion v = u + at

u = 20 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 20 - 1 x t

t = 20 s.

Equation of motion s = ut + 0.5at²

u = 20 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 20 x 20 - 0.5 x 1 x 20² = 200 m

So red train travel 200 m before coming to stop.

For green train:-

Equation of motion v = u + at

u = 40 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 40 - 1 x t

t = 40 s.

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 40 s

a = -1 m/s²

Substituting

s = 40 x 40 - 0.5 x 1 x 40² = 800 m

So green train travel 800 m before coming to stop.

Total distance traveled = 800 + 200 = 1000 m>950 m.

So both trains collide.

Distance traveled by green train when red train stops(t=20s)

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 40 x 20 - 0.5 x 1 x 20² = 600 m

Total distance after 20 s = 600 + 200 = 800 m< 950m . So they collide after red train stops.

Speed of red train when they collide = 0 m/s.

Distance traveled by green train when they collide = 950 - 200 = 750 m

Equation of motion v² = u² + 2as

u = 40 m/s

s= 750 m

a = -1 m/s²

Substituting

v² = 40² - 2 x 1 x 750 = 100

v = 10 m/s

Speed of green train when they collide = 10 m/s.

6 0

3 years ago