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OverLord2011 [107]
2 years ago
15

Two examples of a stopping motion

Physics
2 answers:
Blababa [14]2 years ago
7 0
Add the ball going at a fast speed then slowing down and coming to a complete stop.

Then you will have a car going down a slope the the speed of the car will slow down to the minimum speed of 0.

So these two will be used as examples of stopping motion 
Anton [14]2 years ago
5 0
Pause?Freeze?
Stop?
Halt?
You might be interested in
A diffraction grating, ruled with 300 lines per mm, is illuminated with a white light source at normal incidence.
Vera_Pavlovna [14]

the expression for diffraction grating allows to find the results for the questions for the angular separation are:

i) The third order is Δθ = 0.203 rad.

ii) The first order with water is Δθ = 0.046 rad.

The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.

          d sin θ = m λ

Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.

i) Let's start by looking for the separation between two lines

Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.

         d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm

         d = 3.333 10⁻⁶ m

Let's find the angle of diffraction for the third order (m = 3) for each wavelength.

λ₁ = 400 nm = 400 10⁻⁹ m

         sin θ₁ = \frac{m \ \lambda }{d}m λ/ d

         sin θ₁ = \frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }  

         θ₁ = sin⁻¹ 0.3600

         θ₁ = 0.368 rad

λ₂ = 600 nm = 600 10⁻⁹ m

         sin θ₂ = \frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }  

         θ₂ = sin⁻¹ 0.5401

         θ₂ = 0.571 rad

The angular separation is

         Δθ = θ₂ - θ₁

         Δθ = 0.571 - 0.368

         Δθ = 0.203 rad

ii) In this case, the separation between the network and the observation screen is filled with water.

When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.

           n_i \ sin \theta_1 = n_r \ sin \theta_r

The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.

Let's start looking for the incident angles for the first order of diffraction.

      m = 1

λ₁ = 400 nm

         θ₁ = sin⁻¹  \frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}

         θ₁ = 0.120 rad

λ₂ = 600 nm

        θ₂ = sin⁻¹¹ \frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}

        θ₂ = 0.181 rad

we use the equation of refraction.

         \theta_r  = sin⁻¹ (\frac{n_i}{n_r} \ sin \ \theta_i )

λ₁ = 400 nm  

       θ₁ = sin¹ (\frac{1 sin 0.120}{1.33}

       θ₁ = 0.090 rad

λ₂ = 600 nm

        θ₂ =sin⁻¹  \frac{1 sin 0.181}{1.33}

        θ₂ = 0.1358 rad

The angular separation is

          Δθ = 0.1358 - 0.090

          Δθ = 0.046 rad.

In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:

       i) The third order is Δθ = 0.203 rad.

      ii) The first order with water is Δθ = 0.046 rad.

Learn more here: brainly.com/question/473160

6 0
2 years ago
The plates of a parallel-plate capacitor are oppositely charged and attract each other. Find the expression for the force one pl
Mamont248 [21]

The force exerted by one plate for two parallel plates arrangement is F = QE/2.

<h3>Force one plate exerts on the other</h3>

The force exerted by one plate for two parallel plates arrangement is given as follows;

F = QE/2

where;

  • Q is the charge on one plates
  • E is the electric field due to the charge
  • F is force exerted by one plate

Thus, the force exerted by one plate for two parallel plates arrangement is F = QE/2.

Learn more about force between parallel plates here: brainly.com/question/13590045

5 0
2 years ago
Consider a point 0.5 m above the midpoint of the two charges. As you can verify by removing one of the positive charges, the ele
yKpoI14uk [10]

Answer:

the magnitude of the total electric field is 25 V/m

Explanation:

Given data

point above = 0.5 m

charges = 18 V/m

to find out

the magnitude of the total electric field

solution

the magnitude of the total electric field is 25 V/m because number is less than    the twice magnitude of field by every charge and  the horizontal component in electric field by every charge cancel out and vertical component in field added together

6 0
3 years ago
A piece of aluminum has a volume of 4.89 x 10-3 m3. The coefficient of volume expansion for aluminum is = 69 x 10-6(C°)-1. The t
Leno4ka [110]

Answer:

11.515 Joule

Explanation:

Volume of aluminium = V = 4.89×10⁻³ m³

Coefficient of volume expansion for aluminum = α = 69×10⁻⁶ /°C

Initial temperature = 19.1°C

Final temperature = 357°C

Pressure of air = 1.01×10⁵ Pa

Change in temperature = ΔT= 357-19.1 = 337.9 °C

Change in volume

ΔV = αVΔT

⇒ΔV = 69×10⁻⁶×4.89×10⁻³×337.9

⇒ΔV = 114010.839×10⁻⁹ m³

Work done

W = PΔV

⇒W = 1.01×10⁵×114010.839×10⁻⁹

⇒W = 11.515 J

∴ Work is done by the expanding aluminum is 11.515 Joule

7 0
3 years ago
A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if t
OverLord2011 [107]

Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

m_1u_1 + m_2u_2 = m_1v_1+m_2v_2

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

(97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t =  0.3\ s

Now, the distance covered by the player in this time will be:

s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)

<u>s₁ = 0.022 m</u>

4 0
2 years ago
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