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Arada [10]
3 years ago
9

Two forces are applied to a 17 kg box, as shown. The box is on a smooth surface. Which statement best describes the acceleration

of the box? A. The box accelerates at 1.0 m/s2 to the right because the net force is 17 N to the right. B. The box accelerates at 1.9 m/s2 to the right because the greater force is to the right. C. The box accelerates at 3.0 m/s2 because the combined forces cause the box to accelerate. D. The box does not accelerate, because neither force is large enough to move the box.

Physics
1 answer:
RoseWind [281]3 years ago
7 0
To the picture the answer is A. I can’t answer the typed question because I need the picture for the box
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One problem with weight training as a way to improve overall health is that the results of a weight-training program are not measurable.

B.False

5 0
3 years ago
Read 2 more answers
What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular
monitta

Answer:

a_total = 2 √ (α² + w⁴) ,   a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

          a_total² = a_T²2 + a_{c}²

where

the centripetal acceleration is

  a_{c} = v² / r = w r²

tangential acceleration

   a_T = dv / dt

angular and linear acceleration are related

         a_T = α  r

we substitute in the first equation

       a_total = √ [(α r)² + (w r² )²]

       a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

        w = w₀ + α t

        w = 0 + 1.0 2

        w = 2.0rad / s

       

we substitute

        a_total = r √(1² + 2²) = r √5

        a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

         a_total = 2,236 m

7 0
3 years ago
A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)<br>​
Vadim26 [7]

Answer:

\huge\boxed{\sf f=0.55 \ Hz}

Explanation:

<u>Given Data:</u>

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

<u>Required:</u>

Frequency = f = ?

<u>Formula:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }

<u>Solution:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}

7 0
2 years ago
What is the angular momentum of a 0.25 kg mass rotating on the end of a piece of
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L = r x p = rmv = mr²ω

L = 0.25 x 0.75² x 12.5 = 1.758

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2 years ago
Why is velocity and not just speed important to these pilots
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Velocity is a vector. Therefore, it depends on the direction. Pilots need to know the direction of wind, not just the speed. If the pilot is going South, and there's 5 mph wind going South, they'll be happy, but if the wind is going 5 mph North, they'll be going against the wind.
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3 years ago
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