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3 years ago

What would be the force required to accelerate an object that weighs 35 N up at a rate of .60 m/s^2?

Physics

1 answer:

natta225 [31]3 years ago

3 0

Answer: $2.14\ N$

Explanation:

Given

Force required to accelerate the object at the rate of $0.6\ m/s^2$

The weight of the object is $W=35\ N$

Mass of the object can be identified by dividing the weight by acceleration due to gravity

$\Rightarrow m=\dfrac{W}{g}\\\\\Rightarrow m=\dfrac{35}{9.8}\\\\\Rightarrow m=3.57\ kg$

$\therefore \text{Force F=}m\times a\\\Rightarrow F=3.57\times 0.6\\\Rightarrow F=2.14\ N$

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zepelin [54]

Answer:

Area=1.5(1.5)=2.25m^2

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\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}

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Force

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}

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\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}

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3 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

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To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

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$D = \frac{1.22 (539nm)}{5.11 mm}$

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The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

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3 years ago

Why can't there be a number lower than absolute zero

nexus9112 [7]

Absolute zero is not about numbers. It's about temperature, and the
motion of molecules in gases.

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The faster they're all moving, the warmer the substance feels to us.
The slower they're all moving, the cooler the substance feels to us.

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3 years ago

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A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

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2 years ago

What were some major reasons for the american television from analog broadcast to digital broadcast?

Vika [28.1K]

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Explanation:

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3 years ago

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