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3 years ago

What are shock waves? How are they produced?

1 answer:

7 0

This phenomenon is known as a shock wave. Shock waves are also produced if the aircraft moves faster than the speed of sound. If a moving source of sound moves faster than sound, the source will always be ahead of the waves that it produces.

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Lourdes mixes several ingredients in a bowl, creating a cake batter. She holds the bowl up and turns it upside down, causing the

sasho [114]

friction and gravity ....................

4 0

3 years ago

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi

True [87]

Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

H = maximum difference in height of the track = 19 m

g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

The gravitational potential energy of the coaster on the top of the 11 m high hill (relative to its initial starting point) is:

PEg = m g h

PEg = (410 kg) (9.8 m/s²) (11 m)

PEg = 44198 J

To reach that height, the elastic potential energy stored in the spring must be the same, so:

PEg = PEe = k d² / 2

(44198 J) = k (2.3 m)² / 2

k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

K = 19045 N/m - answer spring constant

When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

PEg = 26656 J

The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

TE = KE = m V² / 2

(77982 J) = (340kg) V² / 2

V = 21.46 m/s - answer maximum speed

4 0

3 years ago

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t

dybincka [34]

Answer:

A) t = 7.0 s

B) x = 25 m

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1 m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0

4 years ago

A man exerts a constant force to pull a 51-kg box across a floor at constant speed. He exerts this force by attaching a rope to

Vitek1552 [10]

Answer:

W=561.41 J

Explanation:

Given that

m = 51 kg

μk = 0.12

θ = 36.9∘

Lets F is the force applied by man

Given that block is moving at constant speed it mans that acceleration is zero.

Horizontal force = F cos θ

Vertical force = F sinθ

Friction force Fr= μk N

N + F sinθ = m g

N = m g - F sinθ

Fr = μk (m g - F sinθ)

For equilibrium

F cos θ = μk (m g - F sinθ)

F ( cos θ +μk sinθ) = μk (m g

Now by putting the values

F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10

F= 70.2 N

We know that Work

W= F cos θ .d

W= 70.2 x cos 36.9∘ x 10

W=561.41 J

3 0

4 years ago

The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of

Ahat [919]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is $=4.51m/s$

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

Now to obtain the initial energy stored

Let U denote the initial energy stored and

$U = \frac{1}{2} kx^2$

Where x is the length the spring is displaced

k is the force constant of the string

$U = \frac{1}{2} * 627 * (0.6)^2$

$= 112.86 J$

Now referring to the formula above

i.e Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

$\frac{1}{2} mv^2 = 112.86 - \mu_kmgx$

$v^2 = \frac{2(112.86 -\mu_kmgx)}{m}$

$v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}$

and we are told that coefficient of friction = 0.4 and the mass is 9 kg ,the acceleration due to gravity $= 9.8m/s^2$ this displacement length of spring = 0.6

Therefore $v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }$

$=4.51m/s$

8 0

3 years ago