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The wavelength of red light is 7 x 10^-7 meter. Express this value in nanometers

3241004551 [841]

The wavelength of the red light in "nanometer" is 7× $10^{2}$

Wavelength is given as : 7× $10^{-7}$ meter

1 nanometer = ( $10^{-9}$ meter)

Let X= value of the wavelength in nanometer.

1 nanometer = $10^{-9}$ meter

X nanometer = 7× $10^{-7}$ meter

<em>If we Cross multiply</em>

X nanometer = ( $\frac{ 7* 10^{-7} }{ 10^{-9} }$ )

X= 7× $10^{2}$ nanometer

Therefore, the wavelength in "nanometer" is 7× $10^{2}$

Learn more at :brainly.com/question/12924624?referrer=searchResults

6 0

2 years ago

Any push or pull on an object is Question 5 options: Acceleration Velocity Force Inertia

Kobotan [32]

This would be force. Acceleration is increasing the speed in an object and velocity is how fast an object is going. Also, inertia basically says that an object will stay at rest or in motion unless an outside force acts on it. So, for example, a ball will stay in the air unless gravity acts on it and pulls it down. By definition, force is any action, unopposed (or by itself without any other forces that would do the opposite) will change the motion of an object, so this definitely makes the most sense for the question. Hope this helps!

8 0

3 years ago

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The plane of a rectangular coil, 3.7 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil

gizmo_the_mogwai [7]

Answer:3.77 T/s

Explanation:

Given

Area of cross-section $=3.7\times 3.7 cm^2=13.69 cm^2$

N=no of turns=61

Resistance= $R=8 \Omega$

current (i)=0.04 A

emf induced $=i\times R=0.04\times 8=0.32 V$

emf induced $e=NA\frac{\mathrm{d} B}{\mathrm{d} t}$

$0.32=62\times 13.69\times 10^{-4}\times \frac{\mathrm{d} B}{\mathrm{d} t}$

$\frac{\mathrm{d} B}{\mathrm{d} t}=\frac{0.32\times 10^4}{848.78}=3.77 T/s$

8 0

3 years ago

A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting

ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

$T=\dfrac{38}{8}\\\\T=4.75\ s$

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

k = spring stiffness constant

So,

$k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m$

When the cord is in air,

mg=kx

x = the extension in the cord

$x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m$

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0

2 years ago

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re

spayn [35]

The magnitude of the electric field can be calculated using the equation

$E = \frac{F}{q}$ , where F is the Force acting on the proton and q is the charge of the proton.

We know that the charge of the proton is $q = 1.6 x 10^{-19} C$

We have to calculate the Force first.

We know that F = ma from Newton's 2nd law, where m is the mass of the proton and a is the acceleration.

We know that the mass of the proton $m = (1.67) X 10^{-27} kg$

So it turns out that we have to calculate acceleration before anything else.

In order to calculate the acceleration, we make use of the following data from the question:

Initial Velocity of the proton $V_{i} = 0$

Distance traveled $D = 1.70 cm$ = 0.017 m

Time taken for the travel between the plates $t = (1.48) X 10^{-6} s$

Acceleration a = ?

Using the equation, $D = V_{i}t + \frac{1}{2} at^{2}$ , we get

Knowing that initial velocity is 0, the equation reduces to $D = \frac{1}{2}at^{2}$

Rearranging the equation so as to make a the subject of the formula, we have

$a = \frac{2D}{t^{2} }$

Plugging in the numbers and simplifying gives us a = 1.5 x $10^{10} m/s^{2}$

We can now calculate the Force using F = ma

Plugging in the known values, we get F = 2.5 x $10^{-17} N$

Using this, we can calculate E through the equation $E = \frac{F}{q}$

Plugging the numbers and simplifying gets us E = 156.25 N/C

Thus, the magnitude of the electric field between the plates of the capacitor is 156.25 N/C

B) To calculate the Final Velocity of the proton, we can make use of the equation

$V_{f} = V_{i} + at$

Plugging the numbers in and simplifying gets us $V_{f} = (2.22) * 10^{4} m/s$

5 0

3 years ago

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