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3 years ago

HELP PLEASE!!!!! pulleys, wheels, and axels.

Physics

2 answers:

Sergeu [11.5K]3 years ago

8 0

Your answer is C. 500 N

Hope this helps!!

babymother [125]3 years ago

6 0

The answer is C . 500 N

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According to Newton's second law of motion what is force equal to

lubasha [3.4K]

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the netforce, and inversely proportional to the mass of the object

8 0

3 years ago

Read 2 more answers

CAN ANY OF YALL ANSWER DIS PLS!!!!!!!! I AM GIVING 20 POINTS BTW!!!!!!!

Misha Larkins [42]

Answer:

chemical changes

burning wood

rotting banana

fireworks

physical changes

shredding paper

melting ice

chopping wood

Explanation:

5 0

3 years ago

A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l

baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

$V = \dfrac{\pi r^{2} h}{3}$

The volume of the object is therefore;

$\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3$

Where 6 cm is above the oil level we have;

$\dfrac{\pi \times \left (6 \times \dfrac{4}{14} \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3$ above the oil level

Therefore, volume of the oil displaced = $68\tfrac{116}{147}\pi$ cm³ = 216.11 cm³

The density of the object is thus;

$\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times Density \ of \ the \ oil = \dfrac{316}{343} \right ) \times Density \ of \ the \ oil$

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = $2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3$

The fraction of the volume displaced, x, after immersing the object is given as follows;

$x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461$

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

8 0

3 years ago

Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field

kirza4 [7]

Answer: the work done by the force is 0

Explanation:

F (x², xy)

121 = 11²

so R = x² + y² = 11²

p = x². Q = xy

Δp/Δy = 0, ΔQ/Δx

using Green's theorem

woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA

= (x² + y² = 121)_∫∫ yΔA

now let x = rcosФ, y = rsinФ

ΔA = rΔrΔФ

so r from 0 to 11

and Ф from 0 to 2π

= 0_∫^2π 0_∫^11 rsinФ × rΔrΔФ

= 0_∫^2π SinФΔФ 0_∫^11 r²Δr

= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0

therefore the work done by the force is 0

3 0

3 years ago

Please help me! I don’t understand these types of problems.

alina1380 [7]

Answer:

60 m

Explanation:

also if u ever get stuck just look up displacment caculator it can help . Please give me brainlist

8 0

2 years ago