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Zolol [24]
3 years ago
9

Convert 81.9 km into meters.

Physics
1 answer:
RUDIKE [14]3 years ago
3 0

Answer: 81900

Explanation: 81.9 x 1000 = 81900.

King Henry Doesn’t Usually Drink Chocolate Milk

                                  Meters

(Meters is Usually) (U)

Goes right 3 times so 3 0's so 1000.

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How much meters is a mile
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Roughly 1609 meters in one mile

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3 years ago
a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T
wlad13 [49]

Explanation:

Given:F=m\ddot{x}=Fe^{-\frac{t}{T}}

Solving for \ddot{x}:

\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}

where:

T=\sqrt{\frac{m}{F}}

Integrating to get \dot{x} with initial conditions \dot{x}(0)=0:

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}

Integrating to get x with initial conditions x(0) = 0:

x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}

When t=T:

x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})

4 0
3 years ago
A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is
Tpy6a [65]

Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

  • Force applied = 100 N
  • Distance covered = 5 metres
  • Time = 10 seconds

<h3><u>To find :</u></h3>

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

\bf \boxed{Work = Force \times displacement}

Now, substituting values in the above formula;

Work = 100 × 5

= 500 Nm or 500 J

We know that,

\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}

Substituting values in above formula;

Power = 500/ 10

= 50 Nm/s or 50 W

Hence, power = 50 W .

5 0
3 years ago
Winner of five swimming gold medals at the 1988 olympics
Fed [463]
Matt Biondi..?
(I don’t know if it’s right, sorry if it is wrong)
:)
6 0
3 years ago
Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv
deff fn [24]

Answer:

 electric field E = - k Q (1 /r(r-a)), force    F = - k Q qo / r (r-a) and force for r>>a    F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

       

           E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

          λ = q / x

As it is constant, we can write it based on differentials

         λ = dq / dx

         dq = λ dx

We already have all the terms, let's  integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

         E = k ∫ λ dx / x²

         E = k la (- 1 / x)

Let's get the negative sign from the parentheses

         E = - k λ (1 / x)

         E = - k λ (1 /(r-a)  -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

         E = - k Q/a  [a / r (r-a)]

         E = - k Q (1 /r(r-a))

b) We calculate the force.  

         F = E qo

         F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

         F = - k Qqo / r² (1 -  a/r)

         F ≈ - k Q qo / r²

6 0
3 years ago
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