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3 years ago

What is the minimum force require to move a 5kg wooden crate on a wooden floor?

1 answer:

7 0

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with µ. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ F = p - f = 0

• net vertical:

∑ F = n - w = 0

where

p = magnitude of the pushing force

f = mag. of friction

n = mag. of the normal force

w = weight of the crate

The second equation gives

n = w = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of µ, so

f = µ (49 N) = 49µ N

To overcome static friction, the push has to exceed this in magnitude, so that

p > 49µ N

For instance, if p = 0.25, then p would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

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Eva8 [605]

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4 years ago

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German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δ???

fredd [130]

Answer:

$\Delta x = 5.47 \times 10^{-9} m$

Explanation:

As we know by the principle of uncertainty that the product of uncertainty in position and uncertainty in momentum is given as

$\Delta x \times \Delta P = \frac{h}{4\pi}$

so here we know that

$\Delta v = 0.01 \times 10^6 m/s$

$m = 9.11 \times 10^{-31} kg$

so we have

$\Delta x \times (9.11 \times 10^{-31})(0.01 \times 10^6) = \frac{6.26 \times 10^{-34}}{4\pi}$

$\Delta x = 5.47 \times 10^{-9} m$

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3 years ago

Which of the following is not a constant or control for this experiment?

SCORPION-xisa [38]

D.the amount of time in sun

8 0

3 years ago

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When a Lunar Module landed on the Moon, it used thrusters to slow its descent to the surface. When other spacecraft are returned

mote1985 [20]

A parachute is a device designed to generate a LOT of air resistance.
Parachutes do that very well in places where there's any air to work with.
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3 years ago

The drawing shows a large cube (mass = 28.6 kg) being accelerated across a horizontal frictionless surface by a horizontal force

MrRissso [65]

Answer:

P= 454.11 N

Explanation:

Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

$P= (M+m)*a\\a = \frac{P}{28.6 +4.3}\\a = \frac{P}{32.9}$

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

$F = m * a*\mu \\F= 4.3*0.710*\frac{P}{32.9}\\F=3.053*\frac{P}{32.9}$

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

$3.053*\frac{P}{32.9} = 4.3 * g\\\\P = \frac{4.3*9.8*32.9}{3.053} \\P= 454.11 N$

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N

8 0

3 years ago