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3 years ago

Name two kinds of longitudinal waves and explain how you know they are longitudinal.

1 answer:

8 0

A transverse wave and a longitudinal wave.

Transverse:wave particles move at medium speed in perpendicular to the direction that the waves move

Longitudinal:wave particles move at medium speed in parallel to the direction that the wave moves

Hope this helps ^-^

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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

what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2$

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2$

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2$

Hence, this is the required solution.

8 0

2 years ago

Read 2 more answers

What is the minimum runway length that will serve? hint: you can solve this problem using ratios without having any additional i

ehidna [41]

There are many factors that determine if an aircraft can operate from a given airport. Of course the availability of certain services, such as fuel, access to air stairs and maintenance are all necessary. But before considering anything else, one must determine if the plane can physically land at an airport, and equally as important, take off.

What is the minimum runway length that will serve?

Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.

A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.

Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.

To learn more about runway

brainly.com/question/11553726

#SPJ4

6 0

1 year ago

I need help with questions 6-8. Thank you!! Image is attached

Digiron [165]

6) b) 2.7 m/s

7) b) DCA

8) b) B

Explanation:

In a displacement-time plot, the slope of the line is given by

$m=\frac{\Delta y}{\Delta x}$

where

$\Delta y$ is the change in the y-variable, so it is the displacement

$\Delta x$ is the change in the x-variable, so it is the time elapsed

So, the slope of the line in a displacement-time plot corresponds to the velocity:

$v=m=\frac{d}{t}$

Therefore, to find the velocity of the object, we have to estimate the slope of its curve.

To estimate the velocity of object B, we have to estimate the slope of the line tangent to curve B at 10 seconds.

By doing an estimate by eye, we see that the displacement of object B changes from -10 m to 0 m when time increases from about 8 s to 12 s, so the velocity is about:

$v=\frac{0-(-10)}{12-8}\sim 2.5 m/s$

So the closest option is b) 2.7 m/s.

As we said in part A, the velocity of each object is given by the slope of each curve.

Therefore:

- The steeper the curve, the higher the velocity

- The less steep the curve, the lower the velocity

From the graph, we observe that, among A, C and D:

- Curve D has the largest slope (in absolute value), so object D has the largest magnitude of the velocity

- Curve C is less steep than curve C, so object C has the second largest magnitude of velocity

- Curve A is flat, so the slope is zero, so its velocity is zero

So, from greatest magnitude to lowest magnitude of velocity, we have:

b) DCA

In the graph, the overall displacement of each object is given by the change in the y-variable, $\Delta y$ .

This means that the object with largest displacement is the object whose curve has the largest variation in y.

From the graph, we see that:

- Object b has the largest variation in y, from -15 m to 30 m, so

$\Delta y=30-(-15)=45 m$

- Then, object D has the second largest displacement (in magnitude), from -15 m to 25 m,

$|\Delta y| = 25 -(-15)=40 m$

Finally, object C has displacement

$\Delta y = 20-(-5)=25 m$

While object A has displacement zero. Therefore, the correct option is

b) B

3 0

3 years ago

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

tresset_1 [31]

Answer:

a) Acceleration of the car is given as

$a_{car} = -21 m/s^2$

b) Acceleration of the truck is given as

$a_{truck} = 10.15 m/s^2$

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

now we have

$1400 (6.32) + 2900(0) = (1400 + 2900) v$

$v = 2.06 m/s$

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

$a_{car} = \frac{v_f - v_i}{t}$

$a_{car} = \frac{2.06 - 6.32}{0.203}$

$a_{car} = -21 m/s^2$

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

$a_{truck} = \frac{v_f - v_i}{t}$

$a_{truck} = \frac{2.06 - 0}{0.203}$

$a_{truck} = 10.15 m/s^2$

5 0

2 years ago