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-BARSIC- [3]
3 years ago
11

Which best defines transparent and provides an example of a material that is transparent to light?

Physics
1 answer:
Crank3 years ago
6 0
<span>Answer: option B. The ability to transmit light; for example water. The transparency is due to the pass of the light. The pass of light is what is called transmission of light. Light may be reflected, refracted or transmitted. Remember light is combination of several waves with different frequencies. When light falls upon an object the interaction may be some frequencies are reflected and other transmitted. Transparent objects are those that permit tha pass of (transmit) most light waves and so let the eyes and instruments to see through them.</span>
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Dragsters can actually reach a top speed of 145.0 m/s in only 4.45 s. (a) Calculate the average acceleration for such a dragster
astraxan [27]

Answer:

a) 32.58 m/s²

b) 161.84 m/s

Explanation:

Initial velocity = u = 0

Final velocity = v = 145 m/s

Time taken = t = 4.45 s

s = Displacement of dragster = 402 m

a = Acceleration

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{145-0}{4.45}\\\Rightarrow a=32.58\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 32.58\times 402-0^2}\\\Rightarrow v=161.84\ m/s

The final velocity is greater than the velocity used to find the average acceleration due to the gear changes. The first gear in a dragster has the most amount of toque which means the acceleration will be maximum. The final gears have less torque which means the acceleration is lower here. The final gears have less acceleration but can spin faster which makes the dragster able to reach higher speeds but slowly.

7 0
3 years ago
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3 years ago
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The units of work,energy and power are...............units​
Andreas93 [3]

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I need help with questions 6-8. Thank you!! Image is attached
Digiron [165]

6) b) 2.7 m/s

7) b) DCA

8) b) B

Explanation:

6)

In a displacement-time plot, the slope of the line is given by

m=\frac{\Delta y}{\Delta x}

where

\Delta y is the change in the y-variable, so it is the displacement

\Delta x is the change in the x-variable, so it is the time elapsed

So, the slope of the line in a displacement-time plot corresponds to the velocity:

v=m=\frac{d}{t}

Therefore, to find the velocity of the object, we have to estimate the slope of its curve.

To estimate the velocity of object B, we have to estimate the slope of the line tangent to curve B at 10 seconds.

By doing an estimate by eye, we see that the displacement of object B changes from -10 m to 0 m when time increases from about 8 s to 12 s, so the velocity is about:

v=\frac{0-(-10)}{12-8}\sim 2.5 m/s

So the closest option is b) 2.7 m/s.

7)

As we said in part A, the velocity of each object is given by the slope of each curve.

Therefore:

- The steeper the curve, the higher the velocity

- The less steep the curve, the lower the velocity

From the graph, we observe that, among A, C and D:

- Curve D has the largest slope (in absolute value), so object D has the largest magnitude of the velocity

- Curve C is less steep than curve C, so object C has the second largest magnitude of velocity

- Curve A is flat, so the slope is zero, so its velocity is zero

So, from greatest magnitude to lowest magnitude of velocity, we have:

b) DCA

8)

In the graph, the overall displacement of each object is given by the change in the y-variable, \Delta y.

This means that the object with largest displacement is the object whose curve has the largest variation in y.

From the graph, we see that:

- Object b has the largest variation in y,  from -15 m to 30 m, so

\Delta y=30-(-15)=45 m

- Then, object D has the second largest displacement (in magnitude), from -15 m to 25 m,

|\Delta y| = 25 -(-15)=40 m

Finally, object C has displacement

\Delta y = 20-(-5)=25 m

While object A has displacement zero. Therefore, the correct option is

b) B

3 0
3 years ago
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passe
never [62]

Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.  

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

  • ∑Fᵧ = maᵧ  
  • T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

  • T₁ = m₂aᵧ + m₂g
  • T₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

  • ∑Fₓ = maₓ
  • F_f - F_g sinΘ = maₓ

The normal force is equal to the x-component of the force of gravity.

  • (F_n · μ_k) - m₁g sinΘ = m₁aₓ
  • (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
  • [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
  • [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
  • [2.539595871] - [-58.0962595] = 6aₓ
  • 60.63585537 = 6aₓ
  • aₓ = 10.1059759 m/s²

Now let's go back to this equation:

  • T₁ = m₂(a + g)  

We have 3 known variables and we can solve for the tension force.

  • T = 2(10.1059759 + 9.8)
  • T = 2(19.9059759)
  • T = 39.8119518 N

The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.

3 0
2 years ago
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