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andrew-mc [135]
3 years ago
8

A diffraction pattern is created on a screen when blue light is passed through a single slit. Does the central bright maximum in

this pattern become wider or narrower, when the blue light is replaced by red light?
Physics
1 answer:
Alex Ar [27]3 years ago
4 0

Answer:

becomes wider

Explanation:

From;   asin θ= m λ

It was said in the question that in this particular instance  blue light is replaced by red light. The wavelength of blue light is less than that of red light, it then follows that the central bright maximum in this pattern widens when red light is used.

Therefore the diffraction pattern becomes wider.

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A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle
Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  \alpha =2.538 \  rad/s^2

Explanation:

From the question we are told that

  The mass of the wheel is  m =  6.9  kg

   The radius is  r =  0.69 \  m

    The radius of gyration is  k_G = 0.4\  m

    The angle is  \theta = 47^o

    The force which the massless bar is subjected to F = 22.5 \  N

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that  point A is the  center of rotation.

  Generally the moment of  inertia about A is mathematically represented as

     I_a =  I_G + M* r^2

Here I_G is the moment of inertia about G with respect to the radius of gyration  which is mathematically represented as

    I_G =  M *  k_G

=>I_a = k_G*  M + M* r^2

=>I_a =0.4 * 6.9  + 6.9 * 0.69^2

=>I_a =6.045 \  kg \cdot m^2

Generally the torque experienced by the wheel  is mathematically represented as

       \tau =  F *  cos (47)

=>     \tau =  22.5 *  cos (47)

=>     \tau =  15.34 \ kg \cdot m^2 \cdot s^{-2}

Generally this torque is also mathematically represented as

     \tau = I_a * \alpha

=>   15.34  =  6.045 * \alpha

=>   \alpha =2.538 \  rad/s^2

4 0
3 years ago
G (where g=9.8 m/s2). If an object’s mass is m=10. kg, what is its weight?
Semenov [28]
F_{w} =m*g \\ F_{w} =10*9.8=98N
7 0
4 years ago
A 4.80 Kg watermelon is dropped from rest from the roof of an 18.0 m building. Calculate the work done by gravity on the waterme
Nonamiya [84]

Answer:

Work, W = 846.72 Joules

Explanation:

Given that,

Mass of the watermelon, m = 4.8 kg

It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.

W = mgh

W=4.8\ kg\times 9.8\ m/s^2\times 18\ m

W = 846.72 Joules

So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.

7 0
3 years ago
¿Que trabajo realizas cuando subes una bolsa de la compra cuya masa es 4,5 kg desde el suelo hasta una mesa de 90 cm de altura?¿
WITCHER [35]
We are going to do this today
6 0
3 years ago
An X-Ray machine delivers a radiation dose of 5mRem/hr. at 3ft from the machine. How far will the X-Ray technician have to move
kupik [55]

Answer:

4.7ft

Explanation:

Pls see attached file

8 0
3 years ago
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