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ludmilkaskok [199]

3 years ago

11

An average human weighs about 650 N. If each of two average humans could carry 1.0 C of excess charge, one positive and one nega

tive, how far apart would they have to be for the electric attraction between them to equal their [email protected] weight?

1 answer:

bezimeni [28]3 years ago

5 0

Answer:

Distance, r = 3721.04 meters

Explanation:

It is given that,

Charges on both humans, $q_1=q_2=1\ C$

Electric force of attraction between them, F = 650 N

We need to find the separation between two humans. It can be solved using the following formula as :

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

$r=\sqrt{\dfrac{kq^2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times (1)^2}{650}}$

r = 3721.04 meters

So, the distance between the humans is 3721.04 meters. Hence, this is the required solution.

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A student uses a stopwatch to measure the period of the pendulum of the Beverly clock in the corridor. His measurements are: (a)

Westkost [7]

Answer:

Reading is close to (b) 13.44 which is the best estimate of the period

Associated error, $\Delta E =0.178 s$

Given:

$t_{a} = 13.54 s$

$t_{b} = 13.44 s$

$t_{c} = 13.89 s$

$t_{d} = 13.41 s$

$t_{e} = 13.17 s$

$t_{f} = 13.22 s$

Solution:

1.The best estimate of the period can be calculated by the mean of the measurements and the one closest to the mean is the best estimate of the measurement:

$Mean, \bar {x} = \fra{sum of all observations}{No. of observation}$

$Mean, \bar {x} = \frac{t_{a} + t_{b} + t_{c} +t_{d} + t_{e} + t_{f}}{6}$

$Mean, \bar {x} = \frac{13.54 + 13.44 + 13.89 + 13.41 + 13.17 + 13.22}{6}$

$Mean, \bar {x} = 13.445 s$

It is close to 13.44 s

2. Associated error is given by:

$\Delta E_{n} = |measured value - actual value|$

$\Delta E_{n} = |t_{n} - \bar {x}|$

where

n = a, b,......, e

Now,

$\Delta E_{a} = |t_{a} - \bar {x}| = |13.54 - 13.44| = 0.01$

$\Delta E_{b} = |t_{b} - \bar {x}| = |13.44 - 13.44| = 0.00$

$\Delta E_{c} = |t_{c} - \bar {x}| = |13.89 - 13.44| = 0.45$

$\Delta E_{d} = |t_{d} - \bar {x}| = |13.41 - 13.44| = 0.03$

$\Delta E_{e} = |t_{e} - \bar {x}| = |13.17 - 13.44| = 0.027$

$\Delta E_{f} = |t_{f} - \bar {x}| = |13.54 - 13.44| = 0.10$

Mean Absolute Error, $\Delta E = \frac{\Sigma E_{n}}{6}$

$\Delta E = \frac{0.01 + 0.00 + 0.45 + 0.03 +0.027 + 1.10}{6}$

$\Delta E =0.178 s$

3. The assumption behind the estimation is population is considered to distributed normally.

6 0

3 years ago

You are a lifeguard and spot a drowning child 30 meters along the shore and 60 meters from the shore to the child. You run along

stiv31 [10]

Answer:

The lifeguard should run approximately 17.752 meters along the shore, before, jumping in the water

Explanation:

The given parameters are;

The rate at which the lifeguard runs = 5 m/s

The rate at which the lifeguard swims = 1 m/s

The horizontal distance of the child from the lifeguard = 30 meters along the shore

The vertical distance of the child from the lifeguard = 60 meters along the shore

Let x represent the distance the lifeguard runs

We have;

The distance the lifeguard swims = √((30 - x)² + 60²)

Time = Distance/Speed

The time the lifeguard runs = x/5

The time the lifeguard swims = √((30 - x)² + 60²)/1

The total time = √((30 - x)² + 60²) + x/5

The minimum time is given by finding the derivative and equating the result to zero, as follows;

Using an online application, we have;

d(√((30 - x)² + 60²) + x/5)/dx = 1/5 - (30 - x)/(√((30 - x)² + 60²)) = 0

Which gives;

1/5 - (30 - x)/(√(x² - 60·x + 4500) = 0

(30 - x)/(√(x² - 60·x + 4500)) = 1/5

5×(30 - x) = √(x² - 60·x + 4500)

We square both sides to get;

(5×(30 - x))² = (x² - 60·x + 4500)

(5×(30 - x))² - (x² - 60·x + 4500) = 0

25·x² - 1500·x + 22500 - x² + 60·x - 4500 = 0

24·x² - 1440·x + 18000 = 0

Dividing n=by 24 gives;

24/24·x² - 1440/24·x + 18000/24 = 0

x² - 60·x + 750 = 0

By the quadratic formula, we have;

x = (60 ± √((-60)² - 4×1×750))/(2 × 1) =

Using an online application, we have;

x = (60 ± 10·√6)/(2)

x = 30 + 5·√6 or x = 30 - 5·√6

x ≈ 42.25 m and x ≈ 17.752 m

At x = 42.25

Time = √((30 - 42.247)² + 60²) + 42.247/5 ≈ 69.69 seconds

At x = 17.75

Time = √((30 - 17.752)² + 60²) + 17.752/5 ≈ 64.79 seconds

Therefore, the route with the shortest time is when the lifeguard runs approximately 17.752 meters (rounded to three decimal places) along the shore, before, diving in the water

5 0

3 years ago

An airplane is trying to fly at a speed of 300 miles per hour straight North, but a 70 miles per hour wind is blowing toward the

OlgaM077 [116]

Answer:

V = 308.06 miles/hour

θ = 13.13° west of north.

Explanation:

Airplane speed = V₁ = 300 miles/hour

Wind speed = V₂ = 70 miles/hour

Resultant speed of the plane = V = ?

As airplane is trying to fly straight North and wind is blowing toward the west. So the angle between airplane velocity and wind velocity is 90°.

By Pythagoras Theorem

V = $\sqrt{V_{1}^{2}+V_{2}^{2} }$

V = $\sqrt{300^{2}+70^{2} }$

V = 308.06 miles/hour

θ = tan⁻¹(70/300)

θ = 13.13° west of north.

3 0

4 years ago

What is SI unit electric current ?

amid [387]

Answer:

the si unit of electric current is Ampere .the flow of charge in a close circuit is called electric current

4 0

3 years ago

Identify a true statement about flow separation. Multiple Choice A. Flow separation is caused due to high temperature gradient i

Helga [31]

Answer:

D. Flow separation is caused due to adverse pressure gradient in the flowing fluid.

Explanation:

Flow separation :

When adverse pressure gradient become dominate then phenomenon of flow separation occurs.In the other words when boundary layer is form against the adverse pressure then phenomenon of flow separation occurs.The adverse pressure means a opposing which act in the opposite to the direction of fluid flow.Due to flow separation eddy formation occurs and these eddy leads to increases the losses in the fluid flow.Due to flow separation fluid leaves the solid surface and form eddies.

So the answer is D.

6 0

3 years ago