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stepan [7]
2 years ago
10

A small block slides without friction along a track toward a circular loop. The block has more than enough speed to remain firml

y in contact with the track as it goes around the loop. The magnitude of the block’s acceleration at the top of the loop isa) zero b) greater than zero but less than g c) equal to g d) greater than g

Physics
1 answer:
lapo4ka [179]2 years ago
3 0

Answer:

d) greater than g

Explanation:

You have to apply Newton's Second Law to have an idea about the value of <em>the block's aceleration.</em>

∑F = ma

Where m is the mass, a is the acceleration and F represents all the forces involved.

You have to draw a free body diagram for the block at the top of the loop. (See the attachment)

Applying Newton's Second Law:

y: N+mg=ma

Where N is the <em>normal force</em> (the contact force that the loop's surface applies to the block) and mg is the <em>weight.</em>

Therefore a= \frac{N}{m} + g

So, you can notice that <em>the magnitude of the block's acceleration</em> is greater than the acceleration of gravity because of the effect of the magnitude N/m.

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3 years ago
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

8 0
3 years ago
3. What exerts a greater force on the table of 2 kg book lying flat or a 2 kg book on its
Darina [25.2K]

Answer:

A book on its side exerts a greater force.

Explanation:

Pressure = Force / Area

Assuming that 1kg = 10N

2kg = 20N

Area of book lying flat = 0.3m × 0.2m

                                     = 0.6m²

Pressure of book lying flat = 20N / 0.6m²

                                            = 30Pa (1 s.f.)

Area of book on its side = 0.2m × 0.05m

                                        = 0.01m²

Pressure of book on its side = 20N / 0.01m²

                                               = 2000Pa (1 s.f.)

Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.

5 0
2 years ago
escribe the differences between the terrestrial and jovian planets and classify each of the 8 planets to the proper type
Radda [10]

Answer:

*******

Explanation:

is obviously the correct answer they are both so different yet so alike

6 0
3 years ago
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ElenaW [278]

Answer:

b)25N

Explanation:

F=ma

F=(5kg)(5m/s^2)

F=25N

4 0
2 years ago
Read 2 more answers
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