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3 years ago

How much force is needed to accelerate a 9760 kg airplane at a rate of 4.6 m/s^2

1 answer:

8 0

Assuming air resistance / friction is negligable:

force = mass × acceleration

f = m × a
f = 9760 × 4.6
f = 44896N

f= 45kN (2 sig fig)

I hope this helps. If you have any questions, feel free to ask

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How much would it cost to cover the entire land area of the U.S. in dollar bills?

agasfer [191]

Answer:

$900 trillion

Explanation:

If Alaska is 20% of the contiguous US, then the approximate area of interest is ...

1200 miles × 3000 miles = 3.6×10^6 square miles.

The size of a dollar bill is about ...

(6.5 cm)·(15.5 cm) = 100.75 cm^2

One mile is 160,934.4 cm, so 1 square mile is about ...

1 mi^2 = (160,934.4 cm)^2 ≈ 2.59·10^10 cm^2

The number of dollars of interest is then ...

(3.6 · 10^6 mi^2)(2.59 · 10^10 cm^2)/(100.75 cm^2) ≈ 9.3·10^14

≈ 930 × 10^12 . . . dollars

It would cost about 900 trillion dollars to cover the land area of the US in $1 bills.

7 0

3 years ago

What is the mass of a 100-N crate of delicious candy

Svetach [21]

Answer:

Explanation:

weight ( w ) = 100 N

mass (m) = ?

WE know

g = 10 m/s²

w = m * g

m = w / g

m = 100 / 10

m = 10 kg

hope it will help :)

6 0

3 years ago

A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap

viktelen [127]

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

5 0

3 years ago

Lus

patriot [66]

Answer:

hey but the person at the top is right

7 0

3 years ago

Define one standard meter, one standard kilogram and one second

lidiya [134]

<h3>The metre is the length of the path travelled by light in vacuum during a time interval of 1299 792 458 of a second.</h3>

6 0

3 years ago