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2 years ago

Which statements accurately describe mechanical waves? Check all that apply.

Physics

2 answers:

Lorico [155]2 years ago

4 0

Answer:

-Energy is transferred through vibrating particles

-An ocean wave moving through water is an example of a mechanical wave

-A longitudinal wave is a type of mechanical wave.

-A transverse wave is a type of mechanical wave

Explanation:

stiks02 [169]2 years ago

3 0

Answers that apply include

Energy is transferred through vibrating particles
An ocean wave moving through water is an example of a mechanical wave
A longitudinal wave is a type of mechanical wave.
A transverse wave is a type of mechanical wave

Mechanical waves can't pass through vacuum like electromagnetic waves because t depends on the transfer of energy between particles of matter (matter that has inertia and elasticity). This energy propagates in the same direction as the wave. Another example of mechanical energy is sound. In addition to longitudinal and transverse waves, another type of mechanical wave is surface waves.

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A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?

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I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
$W_x =mg \sin \theta$
where
m is the mass of the car
g is the acceleration of gravity
$\theta$ is the angle of the hill

Substituting numbers into the formula, we find
$W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N$

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2 years ago

You are participating in a NASA traineeship, working with a group planning a new landing on Mars. Your supervisor has come up wi

aivan3 [116]

Answer:

$h=17005.8 km$

Explanation:

Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass $M=6.39\times10^{23} kg$ at a distance r will be:

$F=\frac{GMm}{r^2}$

where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant.

This force is the centripetal force the satellite experiments, so we can write:

$F=ma_{cp}=mr\omega^2=mr(\frac{2\pi}{T})^2=\frac{4\pi^2mr}{T^2}$

Putting all together:

$\frac{GMm}{r^2}=\frac{4\pi^2mr}{T^2}$

which means:

$r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}$

Which for our values is:

$r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km$

Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km , which leaves us with:

$h=r-R=20395.3km-3389.5 km=17005.8 km$

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2 years ago

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed

$\alpha = 0.745 rad/s2 \rightarrow$ The angular acceleration

$r = 4.65 m \rightarrow$ The distance

The relation between the linear velocity and angular velocity is

$v = r\omega$

Where,

r = Radius

$\omega =$ Angular velocity

At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

$a_c = \frac{r^2\omega^2}{r}$

$a_c = r \omega^2$

$a_c = (4.65 )(1.25 rad/s)^2$

$a_c = 7.265625 m/s^2$

Now the tangential acceleration is given as,

$a_t = \alpha r$

Here,

$\alpha =$ Angular acceleration

r = Radius

$\alpha = (0.745)(4.65)$

$\alpha = 3.46425 m/s^2$

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

$|a| = \sqrt{a_c^2+a_t^2}$

$|a| = \sqrt{(7.265625)^2+(3.46425)^2}$

$|a| = 8.049 m/s^2 \approx 8.05 m/s2$

Therefore the correct answer is C.

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2 years ago

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Leni [432]

Answer:

The force of the car engine.

Explanation:

The work- energy theorem states that the work done on an object is equal to the change in its kinetic energy. Its expression is given by :

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Also, W = F.d

$Fd=\dfrac{1}{2}m(v^2-u^2)$

Where

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UkoKoshka [18]

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2 years ago

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