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Angelina_Jolie [31]
2 years ago
3

Suppose a small cannonball weighing 20 pounds is shot vertically upward, with an initial velocity v0 = 260 ft/s. The answer to t

he question "How high does the cannonball go?" depends on whether we take air resistance into account. If air resistance is ignored and the positive direction is upward, then a model for the state of the cannonball is given by d2s/dt2 = −g (equation (12) of Section 1.3). Since ds/dt = v(t) the last differential equation is the same as dv/dt = −g, where we take g = 32 ft/s2. If air resistance is incorporated into the model, it stands to reason that the maximum height attained by the cannonball must be less than if air resistance is ignored. (a) Assume air resistance is proportional to instantaneous velocity. If the positive direction is upward, a model for the state of the cannonball is given by m dv dt = −mg − kv, where m is the mass of the cannonball and k > 0 is a constant of proportionality. Suppose k = 0.0025 and find the velocity v(t) of the cannonball at time t. Use the result obtained in part (a) to determine the height s(f) of the cannonball measured from ground level. Find the maximum height attained by the cannonball. (Round your answer to two decimal places.)
Physics
1 answer:
serious [3.7K]2 years ago
3 0

Answer:

TO INFINTY WARRR AND BEYONDDDDDDDDDDDDDDDDDDDDD

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An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t
marissa [1.9K]
The energy transferred to the spring is given by:
U= \frac{1}{2}kx^2
where 
k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
k=52.9 N/cm = 5290 N/m
x=40.0 cm=0.4 m

so now we can use these data inside the equation ,to find the energy transferred to the spring:
U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J
4 0
2 years ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
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A.) We use the famous equation proposed by Albert Einstein written below:

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Substituting the value:

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b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
8 0
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The environment affects people, but people don't affect the environment.<br><br><br>TRUE OR FALSE
Drupady [299]

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5 0
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