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Finger [1]
2 years ago
8

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,

the rocket engine is shut down and the sled moves with constant velocity v for a time t2. If the total distance traveled by the sled is 18,350 ft and the total time is 90 s, find the following. (a) the times t1 and t2 t1 = s t2 = s (b) the velocity v
Physics
1 answer:
mash [69]2 years ago
3 0

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

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Answer:

B. 59 kg

Explanation:

From the graph you notice that a linear relation in indicated by the line joining the points such that the points on the line represent the data that show a correct relationship in the experiment.

This means that the point outside the line has an error .

This point is the value 59 kg that does not align with other values which are included in the graph.

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A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.
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Answer:

(a) nearsighted

(b) diverging

(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D

Explanation:

(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.

(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)

(c) the absolute value of the focal length (f) is given by the formula:

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3 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

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