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ivanzaharov [21]
3 years ago
14

A 1.5-kilogram cart initially moves at 2.0 meters per second. It is brought to rest by a constant net force in 0.30 second. What

is the magnitude of the net force?
Physics
1 answer:
AnnZ [28]3 years ago
6 0
Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning.

The cart's acceleration is

                               (0 - 2 m/s) / (0.3 sec)

                           = ( -2 / 0.3 ) (m/s²)  =  -(6 and 2/3) m/s² .

Newton's second law of motion says

                             Force = (mass) x (acceleration) .

For this cart:      Force = (1.5 kg) x ( - 6-2/3 m/s²)

                                       = ( - 1.5 x 20/3 ) (kg-m/s²)

<span>                                       =      </span>- 10 newtons .

<span>The force is negative because it acts opposite to the direction </span>
<span>in which the cart is moving, it causes a negative acceleration, </span>
<span>and it eventually stops the cart.</span>
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Answer:

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From the question we are told that

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Generally the net force is given to be FC(force towards center)

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F_N=F_C-F_g

F_N=\frac{mv^2}{R} -mg

Mathematical we can now derive V

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What is the potential energy of a 3kg ball that is on the ground?
ELEN [110]

This is where we have to admit that gravitational potential energy is
one of those things that depends on the "frame of reference", or
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         Potential energy = (mass) x (gravity) x (<em>height</em>).

So you have to specify <em><u>height above what</u></em> .

-- With respect to the ground, the ball has zero potential energy.
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-- With respect to the floor in your basement, the potential energy is

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(If you let go of it, it will gain 88.2 joules of kinetic energy as it falls
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4 years ago
A 55-kg woman is wearing high heels. If each heel has a circular cross-section 6.0 mm in diameter and she puts all her weight on
bogdanovich [222]

Answer:

The pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

Explanation:

Given;

mass of the woman, m = 55 kg

diameter of the circular heel, d = 6.0 mm

radius of the heel, r = 3.0 mm = 0.003 m

Cross-sectional area of the heel is given by;

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A = π(0.003)²

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The pressure exerted by the woman on the floor is given by;

P = F / A

P = W / A

P = 539 / (2.8278 x 10⁻⁵ )

P = 1.9061 x 10⁷ N/m²

Therefore, the pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

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