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bearhunter [10]
2 years ago
15

A construction worker pushes a crate horizontally on a frictionless floor with a net force of 10\, \text 10N, start text, N, end

text for 4.0\,\text m4. 0m4, point, 0, start text, m, end text. How much kinetic energy does the crate gain?
Physics
1 answer:
Strike441 [17]2 years ago
7 0

Answer:

k_f = 40J

Explanation:

The work and energy theorem says that:

W_f =k_f-k_i

where W_f is the work of the force, k_f the final kinetic energy and k_i the initial kinetic energy.

Addittionally, the work of the force is calculate as force multiply by distance and if the crate inittialy is at rest, the initial kinetic energy is zero, so:

Fd = k_f

where F is the force and d the distance. Then, replacing values, we get:

(10N)(4m) = k_f

40J = k_f

it means that the system gain 40J of kinetic energy.

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A piece of wood that floats on water has a mass of 0.0175 kg. A lead weight is tied to the wood, and the apparent mass with the
crimeas [40]

Answer:

Specific gravity is 0.56

Explanation:

We know that

mass of water displaced by the wood is, m1( apparent mass when wood in air and lead is submerged in water) - m2(the apparent mass when wood and lead both are submerged in water)

= 0.0765 - 0.0452 = 0.0313 Kg

So the specific gravity of the wood is, = mass of wood / mass of water displaced by the wood

= 0.0175/0.0313

=0.56

5 0
2 years ago
The Gaia hypothesis is an example of _____
Fofino [41]
A complex entity involving the Earth's biosphere, atmosphere, oceans, and soil; the totality constituting a feedback or cybernetic system which seeks an optimal physical and chemical environment for life on this planet
4 0
2 years ago
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl
tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

5 0
1 year ago
The tip of the fan blade is 0.61 m from the center of the fan. The fan turns at a constant speed and completes 2 rotation every
Goryan [66]

Answer:

What is the centripetal acceleration of the tip of the fan blade?

6.0 m/s2

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53 m/s2

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Answer is 96

Explanation:

5 0
2 years ago
Two climbers are on a mountain. Simon, of mass m, is sitting on a snow covered slope that makes an angle θ with the horizontal.
elena-14-01-66 [18.8K]

Answer:

Explanation:

It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.

Let in equilibrium , tension in rope be T

For balancing Joe

T = M g

For balancing Simon

friction + T = mgsinθ

μmgcosθ+T = mgsinθ

μmgcosθ+Mg = mgsinθ

M = (msinθ - μmcosθ)

M = m(sinθ - μcosθ)

5 0
2 years ago
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