All categories

4 years ago

What wavelength of light would you report in units of nm, if the light had a wavelength of 7.60 x 10-10 m?

Physics

1 answer:

mezya [45]4 years ago

4 0

Explanation:

One meter is equal to $10^{9}nm$ , so:

$7.60*10^{-10}m*\frac{10^9nm}{1m}=0.76nm$

This wavelength does not correspond to any color of the visible spectrum. A wavelength of $0.76nm$ would correspond to soft X-rays, since these electromagnetic radiation have a wavelegth range from $1nm$ to $0.1nm$ .

You might be interested in

The law of reflection states that if the angle of incidence is 19 degrees, the angle of reflection is ___ degrees.

Wewaii [24]

using the law of refraction, the incidence is equal to the reflection, but not refraction

4 0

3 years ago

Read 2 more answers

An attack helicopter is equipped with a 20- mm cannon that fires 88.7 g shells in the forward direction with a muzzle speed of 9

Svetradugi [14.3K]

The impulse imparted to the shells equals the change in the momentum:
Fav*(Delta t)= Delta m*v.
The mass change is
Delta m= n*m= (89.9shells)*(88.7g)=7.97Kg
So the average force is
F=((v)*(Delta m))/t= ((929)*(7.97))/4.84=1529.78 N
Since the velocity of the shells is much greater than the velocity of the helicopter, there is no need to use relative velocity.

4 0

3 years ago

Which of the following units are used to describe acceleration?

Pavel [41]

Answer:

m/s2 m/s

Explanation:

7 0

3 years ago

3. Is the relationship between the gravitational force and the mass directly proportional or

Dmitrij [34]

Answer:

Directly Proportional

Explanation:

Gravitational force can be calculated with the equation F = g(m1 * m2)/ r^2

So if we increase mass, force will also increase because mass is in the numerator.

5 0

3 years ago

MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle

labwork [276]

Answer:

(a) $y_{max}=0.423m$

(b) $\alpha =64.3^{o}$

Explanation:

Given data

$v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle$

Solution

For Part (a)

As the velocity component in direction of y is given by:

$v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s$

The maximum displacement is given by:

$v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m$

For Part (b)

To reach y=46cm =0.46m apply:

$0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}$

5 0

3 years ago

Read 2 more answers