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3 years ago

Summarize: Based on what you have learned, how will the sound that the observer hears

Physics

1 answer:

Reil [10]3 years ago

3 0

Answer: The sound will change due to changes in frequency and the wavelength of the airplane.

Explanation: Let assume that the observer is at a stationary position. The wavelength of the sound from the airplane reduces and the frequency increases as the plane is moving toward the observer. As the airplane passes by, that is, moving away from the observer, the frequency starts to reduce while the wavelength of the sound starts to increase.

The sound that the observer hears will change base on the illustration above.

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18.5 miles per second (30 km/sec). Choose the Earth movement that best relates to this description.

Vladimir79 [104]

Answer:

Earth orbits the SUn

Explanation:

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3 years ago

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How could rescue workers use squeezing or compressing to get energy to their flashlights during rescue missions?

Neko [114]

Answer:

Explanation:

During rescue missions, different types of energy can be devices for flashlight, this could be human powered energy such as squeezing or compressing. In flashlight electrical energy is converted to light and thermal energy.

A squeezing or compressing to get energy for flashlight can be regarded as "DYNAMO PROCESS" it involves spinning of "fly wheels" into the flashlight through consistent squeezing ,which is connected to a dynamo(Dynamo supply electrical current). Hence the needed light is seen on the bulb of the flashlight.

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3 years ago

Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the

mash [69]

Answer:

The value is $} N = 66 \ blocks$

Explanation:

From the question we are told that

The weight of the block is $W_b = 100 \ N$

The dimension of the block is $d = 0.400 m \ X \ 0.250 \ m \ X \ 0.130 \ m$

Generally two atmosphere is equivalent to

$P_{2atm} = 2 * 1.013 *10^{5} = 202600 \ N/m^2$

Generally 1 atm = $1.013 *10^{5} N/m^2$

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

$A = 0.250 * 0.130$

=> $A = 0.0325 \ m^2$

Generally the force due to this blocks is mathematically represented as

$F = N * W_b$

Here N is the number of blocks

$} 202600 = \frac{N * 100 }{ 0.0325}$

=> $} N = 66 \ blocks$

3 0

3 years ago

A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio

Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

$\sum F_{x}=0$

we can see there are only two forces in x, which are the weight on x and the static friction, so:

$-W_{x}+f_{s}=0$

when solving for the static friction we get:

$f_{s}=W_{x}$

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

$W_{x}=mg sin \alpha$

we can substitute this on our sum of forces equation:

$f_{s}=mg sin \alpha$

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

$f_{s}=N\mu_{s}$

so we can substitute this on our equation:

$N\mu_{s}=mg sin \alpha$

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

$\sum F_{y}=0$

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

$N-W_{y}=0$

when solving for N we get:

$N=W_{y}$

When seeing the free-body diagram we can determine that:

$W_{y}=mg cos \alpha$

so we can substitute that in the sum of y-forces equation, so we get:

$N=mg cos \alpha$

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

$mg cos \alpha \mu_{s}=mg sin \alpha$

we can divide both sides of the equation into mg so we get:

$cos \alpha \mu_{s}=sin \alpha$

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for $\mu_{s}$ we get:

$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

and we can substitute the given value so we get:

$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

8 0

3 years ago

A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia

Bingel [31]

Answer

given,

diameter of the pipe is = (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

$F = \dfrac{mv^2}{r}$

equating both the force equation

$mg = \dfrac{mv^2}{r}$

$v = \sqrt{gr}$

r = d/2 = 14/ 2 = 7 ft

r = 4.27/2 = 2.135 m

g = 32 ft/s² or g = 9.8 m/s²

$v = \sqrt{32 \times 7}$

v = 14.96 ft/s

$v = \sqrt{9.8 \times 2.135}$

v = 4.57 m/s

5 0

3 years ago