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2 years ago

As an apple falls from a tree, which types of energy are changing?

1 answer:

6 0

When an apple falls from the tree to the ground, its energy of position (stored as gravitational potential energy) is converted to kinetic energy, the energy of motion, as it falls. When the apple hits the ground, kinetic energy is transformed into heat energy.

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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

2 years ago

Question number 8 <br> Plz help

Dennis_Churaev [7]

To me, that sounds like the "Law of Conservation of Energy".

5 0

2 years ago

Three common elements that can reorient their electrons into magnetic domains and become magnetic are iron, nickel, and ________

ohaa [14]

Answer:

Cobalt

Explanation:

8 0

2 years ago

A certain piece of metal (density 9.45 grams per cubic centimeter) has the shape of a hockey puck with a diameter of 13 cm and a

atroni [7]

Answer:

0.0195 m

Explanation:

$\rho _{p}$ = density of hockey puck = 9.45 gcm⁻³ = 9450 kgm³

$d_{p}$ = diameter of hockey puck = 13 cm = 0.13 m

$h_{p}$ = height of hockey puck = 2.8 cm = 0.028 m

$\rho _{m}$ = density of mercury = 13.6 gcm⁻³ = 13600 kgm³

$d$ = depth of puck below surface of mercury

According to Archimedes principle, the weight of puck is balanced by the weight of mercury displaced by puck

Weight of mercury displaced = Weight of puck

$\rho _{m} (0.25)(\pi d_{p}^{2} d ) g = \rho _{p} (0.25)(\pi d_{p}^{2} h_{p} ) g\\\rho _{m} ( d ) = \rho _{p} ( h_{p} )\\(13600) d = (9450) (0.028)\\d = 0.0195 m$

7 0

2 years ago

How much work is done if I use 5N of force to move an object 2 meters

sweet-ann [11.9K]

It’s 10 joules. W=FD, W=5•2=10

8 0

2 years ago