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Volgvan
3 years ago
7

An archer shoots a 150. gram arrow straight up in the air. (Do not try this at home.) The bow was drawn back 75.0 cm by a 175 N

force. How high does the arrow rise assuming air friction is negligible
Physics
1 answer:
Dmitry [639]3 years ago
4 0

Answer:

The arrow rise 44.64 m high assuming air friction is negligible

Explanation:

According to the law of conservation of energy

Kinetic energy =  Potential energy at highest

\frac{1}{2}kx^2=mgh----1

We know that F=kx

So,k = \frac{F}{x}\\k=\frac{175}{0.75}

Substitute the value in 1

\frac{1}{2}(\frac{175}{0.75})(0.75)^2=0.15 \times 9.8 \times h\\\frac{\frac{1}{2}(\frac{175}{0.75})(0.75)^2}{0.15 \times 9.8}=h\\44.64 = h

Hence  the arrow rise 44.64 m high assuming air friction is negligible

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An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr
notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

       Charge = Current \times Time

As two different current is passing at two different times, the net charge will be the different in current.  So,

        \text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

        V=\frac{k q}{R}

Here k = 9 * 10^9, q is the charge and R is the radius. As q=2 \times 10^{-6} \times t and R =17 cm = 0.17 m, then the voltage will be

        V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

The time is required to find to reach the voltage of 1500 V, so

1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}

So, 14 ms is required to reach the potential of 1500 V.

3 0
3 years ago
A police car chases a speeder along a straight road towards a cliff both vehicles move at 160km/h the siren on the police car pr
natta225 [31]

Answer:

f ’= 97.0 Hz

Explanation:

This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer

in this case the source is the police cases that go to vs = 160 km / h

and the observer is vo = 120 km / h

the relationship of the doppler effect is

          f ’= f₀ (v + v₀ / v- v_{s})

let's reduce the magnitude to the SI system

            v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s

            v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s

we substitute in the equation of the Doppler effect

          f ‘= 100 (330+ 33.33 / 330-44.44)

          f ’= 97.0 Hz

4 0
3 years ago
Galileo’s pendulum theory stated that the time taken to swing through one complete cycle of a pendulum depends on what?
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Answer: it depends on the mass of the pendulum or on the size of the arc through which it swings.

Explanation:

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What energy change takes place when the cell absorbs sunlight
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It is called photosynthesis.

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A patient has an order for a drug to be infused at the rate of 5 mcg/kg/min. A 500 ml bag contains 250 mg of the drug and the pa
enot [183]

The flow rate is 17gtts/min.

<h3>What is the drug infusion rate?</h3>
  • The rate of infusion (or dosing rate) in pharmacokinetics refers to the ideal rate at which a drug should be supplied to achieve a steady state of a fixed dose that has been shown to be therapeutically effective. This rate is not only the rate at which a drug is administered.
  • The infusion volume is divided into drops, which is known as a drip-rate. The Drip Rate formula is as follows: Volume (mL) times time (h) equals drip-rate. A patient must get 1,000 mL of intravenous fluids over the course of eight hours.
  • Infusion rates of 3–4 mg/kg per minute are advised by manufacturers to reduce rate-related adverse effects. Usually, the infusion lasts for several hours. Although not advised, rates exceeding 5 mg/kg per hour may be tolerated by some patients.
  • If no negative reactions occur, the rate may be increased in accordance with the table every 30 minutes up to a maximum rate of 3 ml/kg/hour (not to exceed 150 ml/hour).

To find the flow rate is 17gtts/min:

\frac{18 units}{h} \frac{100ml}{100units} \frac{500ml}{250mg} \frac{1mg}{1000mgc} \frac{20gtts}{ml} =\frac{17gtts}{min}

Therefore, The flow rate is 17gtts/min.

To learn more about infusion rate, refer to:

brainly.com/question/22761958

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3 0
1 year ago
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