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3 years ago

What force causes a 1-kg mass to accelerate at a rate of 1 meter per second each second?

Physics

1 answer:

Zigmanuir [339]3 years ago

8 0

Answer:

$F=1N$

Explanation:

Conceptual analysis

To solve this problem we apply Newton's second law:

The acceleration of an object is proportional to the force F acting on it and inversely proportional to its mass m.

a = F / m

Where,

F = m * a Formula (1)

F: Force in Newtons (N)

m: mass in kg

a: acceleration in m/(s^2)

a = v / t Formula (2)

v: speed in m/s

t: time in seconds (s)

Known information

We know the following data:

m = 1kg

v = 1 m/s

t = 1s

Development of the problem:

In the Formula (2): $a = \frac{\frac{1m}{s}}{1s} = 1 \frac{m}{s^2}$

In the Formula (1): $F=1kg* 1 \frac{m}{s^2}=1N$

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Explanation:

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2 years ago

I need to name 20 gases in physics

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4 years ago

Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the ap

Verizon [17]

Answer:

The answer is " $1.144 \times 10^{-4} \ m$ ".

Explanation:

$w=\frac{2.44 \lambda L}{D}\\\\D=\frac{2.44 \lambda L}{w}\\\\$

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3 years ago

An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the

Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

$(n1)*(sin(theta1))=(n2)*(sin(theta2))$

Then we will find the refraction angle relative to the vertical this way:

$(n1/n2)*(sin(theta1))=sin(theta2)$

$(1/1.33)*(sin(45))=sin(theta2)$

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

$tan(32.12)=(X/30cm)$

$X=(tan(32.12))*(30cm)$

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

$tan(32.12)=(24cm/Y)$

$Y=(24cm/tan(32.12))$

Then, Y=38cm

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Lunna [17]

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