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Ad libitum [116K]
3 years ago
13

Tenemos un muelle sobre el que se ejerce una fuerza de 20 N produciéndose una deformación de 5 cm. Determinar: a) la constante r

ecuperadora del muelle b) la deformación que sufriría si se le aplicara una fuerza de 50 N- c) la fuerza aplicada para que se deformara 2 cm
Physics
1 answer:
zzz [600]3 years ago
4 0

Answer:

I have no clue on what to do I am trying my hardest to figure out what to do but this is frustrating I do not know I am trying I'm trying I'm trying I'm trying I'm trying but does it supposed to rain the advocacy try to answer questions have a question on top of questions what I'm trying to do what I can sorry I do not have an answer for you if I could I would give you an answer but I do not have an answer so I do not know hopefully you can find somebody they can figure out the answer for you but I do not know the answer sorry

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¿como se llama el ciclo completo que tarda una onda?
prohojiy [21]

Answer: Light travel

Explanation:

5 0
3 years ago
A 72 kg skydiver is descending on a parachute. His speed is still increasing at 1.2 m/s2. What are the magnitude and direction o
denis23 [38]

Answer:

Parachute exerts a force of 619.2 N upward

The net force is 86.4 N acting downward

Explanation:

As the gravitational acceleration g = 9.8 m/s2, the parachute help reduces the net acceleration to 1.2m/s. So it must exerted an upward acceleration on the skydiver of

9.8 - 1.2 = 8.6 m/s2

Since the skydiver mass is 72 kg, we can use Newton's 2nd law to calculate the force that causes this acceleration of 8.6

F = ma = 8.6*72 = 619.2 N acting upward

The net force is also the product of net acceleration and mass

= 1.2 * 72 = 86.4 N acting downward

5 0
3 years ago
PLEASE HURRY
Sophie [7]

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

• The net force in the perpendicular direction is

∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0

Solving the second equation for <em>n</em> gives

<em>n</em> = <em>mg</em> cos(21°)

<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)

<em>n</em> ≈ 1.83 N

Then the magnitude of friction is

<em>f</em> = <em>µn</em>

<em>f</em> = 0.25 (1.83 N)

<em>f</em> ≈ 0.457 N

Solve for the acceleration <em>a</em> :

-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

<em>a</em> ≈ -5.80 m/s²

so the block is decelerating with magnitude

<em>a</em> = 5.80 m/s²

down the ramp.

5 0
2 years ago
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,
ICE Princess25 [194]
Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂                (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392        (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0
2 years ago
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the horizontal sp
gulaghasi [49]

Answer:

C

Explanation:

horizintal speed stays same

only vertical speed changes

so H speed will stay 30 m/s

6 0
2 years ago
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