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SIZIF [17.4K]
2 years ago
9

A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total

energy.
Physics
1 answer:
Aloiza [94]2 years ago
6 0

Answer:

a = ω^2 A      formula for max acceleration (ignoring sign)

V = ω A         formula for max velocity

V^2 = ω^2 A^2 = a A   from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

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It would most definitely not be a hill or a cliff. I believe it would be a plain.
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3 years ago
Describe one common critique or criticism of Utilitarianism. To what extent do you agree or disagree with this criticism?
lisabon 2012 [21]
The second most common criticism of utilitarianism is that it is impossible to apply - that happiness (etc) cannot be quantified or measured, that there is no way of calculating a trade-off between intensity and extent, or intensity and probability (etc), or comparing happiness to suffering.
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2 years ago
An airplane is flying from Dallas, Texas to Pensacola, Florida. Flying at maximum velocity, it encounters strong winds moving at
jenyasd209 [6]

Answer:

G. It will take twice as long.

Explanation:

Let's call v the original speed of the plane and d the distance between Dallas and Pensacola. The time the plane originally takes to complete the flight is

t=\frac{d}{v}

In this problem, we are told that the plane encounters wind moving at half of its speed: \frac{v}{2}, in the opposite direction. This means that the new speed of the plane is

v'=v-\frac{v}{2}=\frac{v}{2}

And so, the time the plane takes now to complete the flight is

t'=\frac{d}{v/2}=2\frac{d}{v}=2t

So, the plane takes twice the time as before.

4 0
3 years ago
How much heat is required to heat 2 kg of water from 25°C to 40°C?
Dominik [7]

Answer:

126000 J

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.

From the question,

Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C

Constant: c = 4200 J/kg.°C

Substitute these value into equation 1

Q = 2×4200(40-25)

Q = 2×4200×15

Q = 126000 J

5 0
3 years ago
I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite
e-lub [12.9K]

Answer:

The range of powers is    - 5 \ D \le P \le - 2.667\  D

Explanation:

From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be u = -25 \ cm

          Image distance would be  v =  -15 \ cm

To obtain the focal length we would apply the lens formula which is mathematically represented as

              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

           converting to  meters

               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

                 P = - 2.667 \ D

   

From these parameter we can see that with the glass on that for far  point the

         Object distance would be u_f = - \infty \ cm

          Image distance would be  v_f =  -20  \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

                    \frac{1}{f_f} =  \frac{1}{v_f}  -  \frac{1}{u_f}

substituting values

                  \frac{1}{f} =  \frac{1}{-20}  -  \frac{1}{- \infty}

                 \frac{1}{f} =  \frac{1}{-20}  -  0      

                  f_f =  \frac{20}{1}  \ cm

converting to  meters

                f_f =  - \frac{20}{1}  * \frac{1}{100}

               

Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f_f}

Substituting values

                 P = -  \frac{100}{20}  m

                 P = - 5 \ D

This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

3 0
4 years ago
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