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Deffense [45]
3 years ago
14

What are some limitations to your direct observations of your environment? Give an example of an observation that could be made

without using an instrument, but would be improved by using one. Give an example of an observation that could not be made without the use of an instrument.
Physics
1 answer:
UNO [17]3 years ago
8 0

Answer:

I am afriad I don't think we can answer this one as it is asking you about <em>your </em>enviroment.

Explanation:

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So we want to know what is the purpose of a lanyard attached to a safety switch. So in case the operator falls overboard a safety switch is installed and connected to the operators hand or waist. Which ever is more practical. This safety switch turns off the motor. 
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What is true about energy that is added to a closed system?
Setler [38]

The correct answer is B

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Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd
Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

5 0
3 years ago
The doctor writes a prescription for na heparin 20,000 units in 500 ml n.s. infuse over 8 hours. what is the flow rate in ml/hr?
Ne4ueva [31]
I think this type of equation could be conducted in simple division equation since it does not involve drop rate.

we know that there is 500 ml of substance and should be infused within 8 hours period.

So the flow rate in ml/hr would be: 

500/8 = 62.5 ml/hr                                                                                    
8 0
3 years ago
From the average distance of one of jupiter's moons to jupiter and its orbital period around jupiter, we can determine:
den301095 [7]

Answer: Jupiter's mass

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From Kepler's third law:

T^2=\frac{4\pi^2}{GM}a^3

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If the average distance of one of Jupiter's moons to Jupiter and its orbital period around Jupiter is given then mass of the Jupiter can be found:

\Rightarrow M_J=\frac{4\pi^2}{GT_m^2}a_m^3

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