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3 years ago

Perpendicular parking spaces require turning at a ______-degree angle.

1 answer:

7 0

Perpendicular means at 90 degree angle. so,
<span>Perpendicular parking spaces require turning at a 90 degree angle.

When you are going to park perpendicularly, you need a distance of 7 to 8 feet from the vehicle you are parking next to, and when you are parking parallel, you need 5 feet distance from the vehicle you are parking next to.</span>

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An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part

guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

$E=\dfrac{kq}{r^2}$

k is electrostatic constant

if r = 2 m, electric field is given by :

$E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)$

If r = 1 m, electric field is given by :

$E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)$

Dividing equation (1) and (2) we get :

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1$

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0

3 years ago

What's the frequency of a wave with a wavelength of 10 and velocity of 200m/s?

uranmaximum [27]

Answer:

$\boxed {\boxed {\sf 20 \ Hz}}$

Explanation:

The frequency of a wave can be found using the following formula.

$f=\frac{v}{\lambda}$

where <em>f</em> is the frequency, <em>v</em> is the velocity/wave speed, and λ is the wavelength.

The wavelength is 10 meters and the velocity is 200 meters per second.

1 m/s can also be written as 1 m*s^-1

Therefore:

$v= 200 \ m*s^{-1} \\\lambda = 10 \ m$

Substitute the values into the formula.

$f=\frac{200 \ m*s^{-1}}{10 \ m}$

Divide and note that the meters (m) will cancel each other out.

$f=\frac{200 \ s^{-1}}{10 \ }$

$f=20 \ s^{-1}$

1 s^-1 is equal to Hertz
Therefore, our answer of 20 s^-1 is equal to 20 Hz

$f= 20 \ Hz$

The frequency of the wave is <u>20 Hertz</u>

7 0

2 years ago

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

The orbital speed is $v= 2.6*10^{3} m/s$

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

3 years ago

Two examples of a stopping motion

Anton [14]

Pause?Freeze?
Stop?
Halt?

5 0

3 years ago

Read 2 more answers

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

#SPJ9

3 0

1 year ago