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Contact [7]
3 years ago
8

How are frequency and period related to each other? A. They are the same for any given wave B. They have the same magnitude but

opposite signs C. They are reciprocals of each other D. They are both found by squaring the amplitude of a wave
Physics
1 answer:
BabaBlast [244]3 years ago
3 0
Either of them is. 1/(the other one). That's 'C' .
You might be interested in
Mo is on a baseball team and hears that a ball thrown at a 45 degree angle from the ground will travel the furthest distance. Ho
Galina-37 [17]

Answer:

Explanation:

Usually the angle between the y axis  and x axis is 90° and we know that for furthest travel the degree angle must be 45° with the horizontal, Mo must release the ball about halfway between straight ahead and straight up

3 0
3 years ago
The speed of sound is measured to be 340 m/s on a certain day.
ziro4ka [17]
Answer: 1,224 km/h

Explanation:

To do this, we pick the first unit and convert
Picking m first and converting to km:
Since we're converting from a non-prefix to a prefix, we divide the value by the prefix were taking it to. In this case, kilo = 10³ which means we're going to divide our value by 1000 to convert it from m to km
340 m/s ÷ 1000 = 0.34 km/s
Now, let's convert our seconds to hour:
We'll need to calculate how many hours is equivalent to one second first;
1 hr = 60×60 seconds
X hr = 1 second
*Cross multiply*
1 × 1 = X × 60 × 60
1 = 3,600 X
X = 1 / 3,600
X = 2.778×10⁻⁴ hour
So, in the place of "1 Second", we're going to be inserting 2.778×10⁻⁴ hour instead
0.34 km / s = 0.34 km / 2.778×10⁻⁴ hour
(0.34 / 2.778×10⁻⁴) km/hour
1,224 km/h.
340 m/s = 1,224 km/h
6 0
2 years ago
A wave has the wavelength equal to 396 nm. what must be the frequency of this wave
Anvisha [2.4K]
This assumes that the wave has velocity c (is light).

7 0
3 years ago
Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin
Vedmedyk [2.9K]

Answer:

The correct answer is 2.8*10^{-5}ms^{-1}

Explanation:

The formula for the electron drift speed is given as follows,

u=I/nAq

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is 8.93/63.5 molcm^{-3}. Converting this number to m³ using very elementary unit conversion we get 140384molm^{-3}. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}

if we convert the area from mm³ to m³ we get A=80*10^{-6}m^{2}.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}

which is our final answer.

6 0
3 years ago
At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri
viva [34]

Answer:

v_{f2} =6.5%v_{i1}

Explanation:

Mass of the ball: m_{1} =0.12kg]

Initial velocity of the ball:   v_{i1}

final velocity of the ball: v_{f1} which is -30/100 of v_{i1} =-0.3v_{i1}

Mass of the bottle: m_{2} =2.4kg

Initial velocity of the bottle: v_{i2}=0m/s

final velocity of the bottle: v_{f2} is unknown (to find)

<em>by using conservation momentum, which stated that the initial momentum is equal to the final momentum.</em>

<em />m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}<em />

<em>so since the bottle is at rest firstly, therefore </em>v_{i2} =0<em />

<em />m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}<em />

<em />m_{1} v_{i1}  =m_{1} v_{f1} +m_{2} v_{f2}<em>         </em><em>equation 1</em>

so now substitute v_{f1} into equation 1

m_{1} v_{i1}  =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}

<em />m_{1} v_{i1}  = -0.3m_{1}v_{i1}  +m_{2} v_{f2}<em />

<em>collect the like terms</em>

m_{1} v_{i1}   +0.3m_{1}v_{i1}  =m_{2} v_{f2}

1.3m_{1} v_{i1}   =m_{2} v_{f2}

divide both  side by m_{2}

v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }

Now substitute

v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2}    =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}

v_{f2} =6.5%v_{i1}

<em />

6 0
3 years ago
Read 2 more answers
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