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kondor19780726 [428]
2 years ago
10

A shopper pushes a 5.32 kg grocery cart

Physics
2 answers:
Juli2301 [7.4K]2 years ago
7 0

Answer:

\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

\text { Acceleration }=\frac{\text { force }}{\text { mass }}

Given that,

Mass = 5.32 kg

\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}

x=-28.7^{\circ}

F = 12.7N

Normal force = mg + F sinx,  

“m” being the object's "mass",  

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

<u>Acceleration of the cart</u>:

\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}

\text { Acceleration }=\frac{58.232}{5.32}

\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}

\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}

maxonik [38]2 years ago
7 0

Answer: 0.53

Explanation: Try looking it up on yahoo.

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Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
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1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

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In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

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                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

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                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

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