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soldi70 [24.7K]
3 years ago
5

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.40 kg box that is sitting on the horizontal,

frictionless surface of a frozen lake. F(x) is the only horizontal force on the box. So, If the box is initially at rest at x=0, what is its speed after it has traveled 16.0 m? Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:v=7.24 m/s

Explanation:

Given

F(x)=18-0.530 x

mass m=8.40 kg

box is at rest at x=0

Distance traveled=16 m

mv\frac{\mathrm{d} v}{\mathrm{d} x}=F=\left ( 18-0530x\right )dx

\int_{0}^{v}vdv=\int_{0}^{16}\left ( 18-0.530x\right )dx

\frac{mv^2}{2}=\left ( 18x-0530\frac{x^2}{2}\right )_0^{16}

v^2=\frac{288-67.841\times 2}{8.40}

v=7.24 m/s

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a motorcycle starts from rest covers 200 meter distance in 6 second calculate final velocity and acceleration​
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Explanation:

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200 = 18 a

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v = u + at

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hope it helps you

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The amount of the lighted side of the moon you can see is the same during
attashe74 [19]

<u>Answer:</u>

The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".

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A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration
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7 0
3 years ago
What's the Coulomb's law?
Ulleksa [173]

<span>
In layman's term: </span>like charges don't attract while opposite charges do<span>electrostatic forces between point A( which is charged) and point B (which is also charged) are proportional to the charge of point A and point B. </span><span>there is also something else about this  law that I don't quite remember.</span>

<span>___________________________________________________</span>

<span />Here is the formula:

<span>F = k x Q1 x Q2/d^<span>2</span></span>

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________________________________________________

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4 0
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Read 2 more answers
A ship leaves a harbor and sails at 23.5 ∘ to the north of due west. After traveling 575 km, how far west is the ship from the h
solniwko [45]

Answer:

D = 527.31 Km

Explanation:

given,

angle of ship, θ = 23.5° N of W

distance travel in the direction = 575 Km

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now,

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inserting all the values

D = 575 x cos 23.5°

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D = 527.31 Km

Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km  

6 0
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