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alina1380 [7]
3 years ago
12

A net force of 10.0 N causes an object to accelerate at 2.00m/s^2. What is the mass of the object?

Physics
1 answer:
madam [21]3 years ago
4 0

5.00kg

random = abcdefghijklmnopqrstuvwxyz


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What is the displacement of the student from the bus stop to the mailbox?
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Answer:C

Explanation:

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Which best contrasts Newton's and Einstein's ideas?
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Answer:

Newton believed that mass tells gravity how much force to exert. Einstein believed that mass tells space-time how to curve.

Explanation:

Isaac Newton believed that bodies on earth had a force of gravity pulling them down as a result of their masses.

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This confirms Newton believing that mass tells gravity how much force to exert and Einstein believing that mass tells space-time how to curve.

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There is a 250-m-high cliff at half dome in yosemite national park in california. suppose a boulder breaks loose from the top of
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Part A. For this part, we use two equations for linear motion:

<span>y = y0 + v0 t + 0.5 g t^2                   ---> 1</span>

<span>vf = v0 + g t                                         ---> 2</span>

First we solve for t using equation 1: y0 = 0 (initial point at top), y = 250 m, v0 = 0 (at rest)

250 = 0.5 (9.8) t^2

t = 7.143 s

Now we solve for final velocity vf using equation 2:

vf = g t

vf = 9.8 (7.143)

vf = 70 m/s

 

Part B. First we solve for the time it takes for the sound to reach the tourist.

t(sound) = 250 / 335 = 0.746 s

Therefore the total time would be:

t = 0.746 s + 0.300 s

t = 1.05 s

 

<span>Hence there is enough time for the tourist to get out before the boulder hits him.</span>

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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat
Sever21 [200]

Answer:

Approximately 325 (rounded down,) assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.

The number of repetitions would increase if efficiency increases.

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Ensure that all quantities involved are in standard units:

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\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}.

Height of the weight (should be in meters, {\rm m}):

\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}.

Energy required to lift the weight by \Delta h = 0.2\; {\rm m} without acceleration:

\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}.

At an efficiency of 0.25, the actual amount of energy required to raise this weight to that height would be:

\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}.

Divide 2.551 \times 10^{5}\; {\rm J} by 784\; {\rm J} to find the number of times this weight could be lifted up within that energy budget:

\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

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