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2 years ago

Which refers to the number of wavelengths that pass a fixed point in a second?

Physics

1 answer:

pav-90 [236]2 years ago

7 0

Answer:

The word that refers to the number of wavelengths that pass a fixed point in a second is:
C.) Frequency

Frequency is also known as hertz.

Explanation:

Have a great rest of your day
#TheWizzer

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The same force is applied to two skateboards. One rolls across the room and the other moves a few feet and comes to a stop. Wher

igor_vitrenko [27]

The longer you spend reading and thinking about this question,
the more defective it appears.

-- In each case, the amount of work done is determined by the strength
of the force AND by the distance the skateboard rolls while you're still 
applying the force. Without some more or different information, the total
distance the skateboard rolls may or may not tell how much work was done
to it.

-- We know that the forces are equal, but we don't know anything about
how far each one rolled while the force continued. All we know is that
one force must have been removed.

-- If one skateboard moves a few feet and comes to a stop, then you
must have stopped pushing it at some time before it stopped, otherwise
it would have kept going.

-- How far did that one roll while you were still pushing it ?

-- Did you also stop pushing the other skateboard at some point, or
did you stick with that one?

-- Did each skateboard both roll the same distance while you continued pushing it ?

I don't think we know enough about the experimental set-up and methods
to decide which skateboard had more work done to it.

8 0

3 years ago

Read 2 more answers

The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point

makvit [3.9K]

To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m

Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.

5 0

3 years ago

Arocket launches at an angle of 33.6 degrees from the horizontal at a

babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, $\theta=33.6$

Initial velocity of the rocket is, $u=58.5$

A vector at an angle $\theta$ with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle $\theta$ with the horizontal, then the horizontal and vertical components are given as:

$A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})$

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

$u_y=u\sin \theta$

Plug in the given values and solve for $u_y$ . This gives,

$u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})$

Therefore, the Y component of initial velocity is 32.37.

4 0

3 years ago

A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a

Aleks [24]

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

3 0

3 years ago

In an old-fashioned amusement park ride, passengers stand inside a 4.9-m-diameter hollow steel cylinder with their backs against

Viefleur [7K]

Answer:

24.07415 rpm

Explanation:

$\mu$ = Coefficient of friction = 0.63

v = Velocity

d = Diameter = 4.9 m

r = Radius = $\frac{d}{2}=\frac{4.9}{2}=2.45\ m$

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

Here the frictional force balances the rider's weight

$f=\mu F_n$

The centripetal force balances the weight of the person

$\mu m\frac{v^2}{r}=mg\\\Rightarrow \mu \frac{v^2}{r}=g\\\Rightarrow v=\sqrt{\frac{gr}{\mu}}\\\Rightarrow v=\sqrt{\frac{9.81\times 2.45}{0.63}}\\\Rightarrow v=6.17656\ m/s$

Velocity is given by

$v=\omega r\\\Rightarrow \omega=\frac{v}{r}\\\Rightarrow \omega=\frac{6.17656}{2.45}\\\Rightarrow \omega=2.52104\ rad/s$

Converting to rpm

$2.52104\times \frac{60}{2\pi}=24.07415\ rpm$

The minimum angular speed for which the ride is safe is 24.07415 rpm

4 0

3 years ago