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vodka [1.7K]
3 years ago
9

Karen has a mass of 51.9 kg as she rides the up escalator at Woodley Park Station of the Washington D.C. Metro. Karen rode a dis

tance of 63.1 m, the longest escalator in the free world. The acceleration of gravity is 9.8 m/s 2 . How much work did the escalator do on Karen if it has an inclination of 26◦ ? Answer in units of J.
Physics
1 answer:
kifflom [539]3 years ago
4 0

Answer:

28852 J

Explanation:

When a force applied in a body produces a displacement in it, the force realized a work. The force that moves Karen is contrary to her weight and must be equal to it.

The work (W) is:

W = F.d.cos(θ), where F is the force, d is the displacement, and θ is the angle.

Knowing that cos(26°) = 0.899, and F = m*g

W = 51.9*9.8*63.1*0.899

W = 28852 J

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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a
Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration=88.6m/s^2

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution=2\pi rad

2.59 rev=2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s

By using \pi=3.14

Angular velocity=\omega=16.27rad/s

Distance from axis=r=0.329 m

Tangential speed=r\omega=16.27\times 0.329=5.4m/s

Radial acceleration=\frac{v^2}{r}

Radial acceleration=\frac{(5.4)^2}{0.329}=88.6m/s^2

6 0
3 years ago
Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0
3 years ago
What is Secular Music's instrument's?​
nasty-shy [4]

Answer:

Drums, harps, recorders, and bagpipes.

Explanation:

8 0
3 years ago
A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca
igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

f=mg\ sin\theta

f=1100\ kg\times 9.8\ m/s^2\ sin(4)

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0
3 years ago
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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