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3 years ago

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2 answers:

tresset_1 [31]3 years ago

6 0

Becomes a +1 ion for this

icang [17]3 years ago

6 0

+1

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HELPPPPP MEEEEE PLEASE I NEED TO SUBMIT IN LESS THAN 10 MINSS

DerKrebs [107]

Answer:

X = 2146.05 m

Explanation:

We need to understand first what is the value we need to calculate here. In this case, we want to know how far from the starting point the package should be released. This is the distance.

We also know that the plane is flying a certain height with an specific speed. And the distance we need to calculate is the distance in X with the following expression:

X = Vt (1)

However we do not know the time that this distance is covered. This time can be determined because we know the height of the plain. This time is referred to the time of flight. And the time of flight can be calculated with the following expression:

t = √2h/g (2)

Where g is gravity acceleration which is 9.8 m/s². Replacing the data into the expression we have:

t = √(2*2500)/9.8

t = 22.59 s

Now replacing into (1) we have:

X = 95 * 22.59

<h2>X = 2146.05 m</h2>

This is the distance where the package should be released.

Hope this helps

6 0

3 years ago

A parallel-plate vacuum capacitor has 8.60 J of energy stored in it. The separation between the plates is 3.80 mm . If the separ

creativ13 [48]

Answer:

Part a)

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

$E = 2.62(8.60) = 22.5 J$

Explanation:

As we know that the energy of capacitor when it is not connected to potential source is given as

$U = \frac{Q^2}{2C}$

As we know that initial energy is given as

$8.60 = \frac{Q^2}{2C}$

now we know that capacitance of parallel plate capacitor is given as

$C = \frac{\epsilon_0A}{d}$

now the new capacitance when distance is changed from 3.80 mm to 1.45 mm

$C' = \frac{Cd}{d'}$

$C' = \frac{C(3.80)}{1.45}$

$C' = 2.62 C$

Now the new energy of the capacitor is given as

$E = \frac{Q^2}{2(2.62C)}$

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

Now if the voltage difference between the plates of capacitor is given constant

now the energy energy of capacitor is

$U = \frac{1}{2}CV^2$

$8.60 = \frac{1}{2}CV^2$

now when capacitance is changed to new value then new energy is given as

$E = \frac{1}{2}C'V^2$

$E = \frac{1}{2}(2.62C)V^2$

$E = 2.62(8.60) = 22.5 J$

6 0

3 years ago

V02 Zone 90-100% MHR; =BPM

shepuryov [24]

Don’t use those links!!! there scams

3 0

3 years ago

Q7, how the ans is 4.9 not 19.6?

taurus [48]

the answer is c in question 7

5 0

1 year ago

A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati

Ainat [17]

Answer:

The asteroid's acceleration at this point is $2.71\ m/s^2$

Explanation:

The equation that governs the trajectory of asteroid is given by :

$x=6.5t-2.3t^3$

The velocity of asteroid is given by :

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2$

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

$6.5-6.9t^2=0\\\\t=0.971\ s$

Acceleration,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t$

Put t = 0.971 s

$a=-13.8\times 0.197\\\\a=-2.71\ m/s^2$

So, the asteroid's acceleration at this point is $2.71\ m/s^2$ and it is decelerating.

6 0

3 years ago