Ask question

Ask question

All categories

2 years ago

6

Tapping the surface of a pan of water generates 17.5 waves per second. If the wavelength of each wave is 45 cm, what is the spee

d of the waves?

1 answer:

ElenaW [278]2 years ago

5 0

Answer:

Speed of the wave is 7.87 m/s.

Explanation:

It is given that, tapping the surface of a pan of water generates 17.5 waves per second.

We know that the number of waves per second is called the frequency of a wave.

So, f = 17.5 Hz

Wavelength of each wave, $\lambda=45\ cm=0.45\ m$

Speed of the wave is given by :

$v=f\lambda$

$v=17.5\times 0.45$

v = 7.87 m/s

So, the speed of the wave is 7.87 m/s. Hence, this is the required solution.

You might be interested in

Floor lamps usually have a base with large inertia, while the long body and top have much less inertia. Part A If you want to sh

Novay_Z [31]

Answer:

When a an object is been rotated its resistance capacity to that rotational force is know as rotational inertia and this mathematically given as

$I = mr^2$

Where m is the mass

r is the rotation radius

For the spinning of the lamp as a baton to work the location of the center of mass of the floor lamp needs to be located

This is more likely to be located closer to base of the lamp as compared to the top, so success of spinning a floor lamp like a baton is highly likely if the lamp is grabbed closer to the base because that is where the position of its center of mass is likely to be.

Explanation:

5 0

2 years ago

An object’s velocity is measured to be vx(t) = α - βt2, where α = 4.00 m/s and β = 2.00 m/s3. At t = 0 the object is at x = 0. (

Leona [35]

Answer:

Explanation:

Given

$v_x(t)=\alpha -\beta t^2$

$\alpha =4\ m/s$

$\beta =2\ m/s^3$

$v_x(t)=4-2t^2$

$v=\frac{\mathrm{d} x}{\mathrm{d} t}$

$\int dx=\int \left ( 4-2t^2\right )dt$

$x=4t-\frac{2}{3}t^3$

acceleration of object

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=-4t$

(b)For maximum positive displacement velocity must be zero at that instant

i.e. $v=0$

$4-2t^2=0$

$t=\pm \sqrt{2}$

substitute the value of t

$x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}$

$x=3.77\ m$

7 0

1 year ago

What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria

Likurg_2 [28]

Answer:

Approximately $\rm 11.2 \; N \cdot kg^{-1}$ at that distance from the center of the planet.

Option A) The low value of $g$ near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of $g$ on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

$\displaystyle \frac{G \cdot M \cdot m}{R^2}$ ,

where

$G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2$ .
$M$ is the mass of the planet, and
$m$ is the mass of the object.

To find an equation for $g$ , divide the equation for gravity by the mass of the object:

$\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}$ .

In this case,

$M = 5.68\times 10^{26}\; \rm kg$ , and
$R = 5.82 \times 10^7\; \rm m$ .

Calculate $g$ based on these values:

$\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}$ .

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for $g$ :

$\displaystyle g = \frac{G \cdot M}{R^2}$ .

The mass of the planet is in the numerator. If two planets are of the same size, $g$ would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since $R$ is in the denominator of $g$ , increasing the value of $R$ while keeping $M$ constant would reduce the value of $g$ . That explains why the value of $g$ near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately $9.81\; \rm N \cdot kg^{-1}$ .

As a side note, $5.82\times 10^7\rm \; m$ likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of $g$ near the cloud tops of Saturn is approximately $\rm 11.2 \; N \cdot kg^{-1}$ .

6 0

2 years ago

Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

matrenka [14]

<span> (a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span> $F=qE$
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
</span> $F=qvB \sin \theta$
where $\theta$ is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
</span> $F=qE$
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
$F=qvB \sin \theta$
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span> $F=qE$
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span> $F=qvB \sin \theta$
<span>is different from zero because v is different from zero.</span>

6 0

2 years ago

Consider a point on a bicycle wheel as the wheel makes exactly four complete revolutions about a fixed axis. Compare the linear

dezoksy [38]

Answer:

Explanation:

Wheel completes four revolution.

The linear displacement is zero.

The angular displacement is 4 x 2π = 8π radian.

So, option (c) is correct.

4 0

2 years ago