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Alborosie
3 years ago
12

A 3.0-kg and a 1.0-kg box rest side-by-side on a smooth, level floor. A horizontal force of 32 N is applied to the 1.0-kg box pu

shing it against the 3.0-kg box, and, as a result, both boxes slide along the floor. How hard do the two boxes push against each other
Physics
1 answer:
stira [4]3 years ago
5 0

Considering both boxes as one body, it would have a total mass of 4.0 kg. By Newton's second law, the 32 N force applies an acceleration <em>a </em>such that

∑ <em>F</em> = 32 N = (4.0 kg) <em>a</em>   →   <em>a</em> = 8.0 m/s²

and both boxes share this acceleration. (There is no friction, so the given force is the only one involved in the direction of the boxes' motion.)

Now consider just the smaller box. It is feeling the effect of the 32 N push in one direction and, as it comes into contact with the larger box, a normal force that points in the opposite direction. Let <em>n</em> be the magnitude of this normal force; this is what you want to find. By Newton's second law,

∑ <em>F</em> = 32 N - <em>n</em> = (1.0 kg) (8.0 m/s²)

<em>n</em> = 32 N - 8.0 N

<em>n</em> = 24 N

Just to make sure that this is consistent: by Newton's third law, the larger box feels the same force but pointing in the opposite direction. On the smaller box, <em>n</em> opposes the pushing force, so points backward. So from the larger box's perspective, <em>n</em> acts on it in the forward direction. This is the only force acting on the larger box, so Newton's second law gives

∑ <em>F</em> = 24 N = (3.0 kg) (8.0 m/s²)

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Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

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r = 1.30 m

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a

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The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

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4 years ago
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algol13
Rotation- sunset,sunrise,moons movement

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4 0
3 years ago
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The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
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Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

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Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

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4 0
3 years ago
A bald eagle is flying to the left with a speed of 34 meters
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Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U  + at

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t = 3 Sec

V = velocity after 3 sec

V  = 34 + (-8)3

 = 34 - 24

 V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

4 0
3 years ago
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