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Vikentia [17]
3 years ago
7

A mass slider m = 0.200 kg rests on a frictionless horizontal air rail connected to a spring with a force constant k = 5.00 N /

m. You pull the slider, stretching the spring 0.100 m, and then release it from rest. The slider returns to its equilibrium position (x = 0). What speed does it have when x = 0.080 m?

Physics
1 answer:
Step2247 [10]3 years ago
5 0
Va ser 0.0900 yo creo preo que esta respuesta te ayude
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How is displacement different from distance?
Anni [7]

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A bus leaves at 9 am with a group of tourists. They travel 350 km before they stop for lunch. Then they travel an additional 250
Anettt [7]

Average speed = Distance traveled / time taken

In this case Time taken = Difference in hours between 3 PM and 9 AM

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8 0
3 years ago
Please Help!!
kotegsom [21]

Answer:

Potential energy of spring = 24 Joules.

Explanation:

Given the following data;

Spring constant = 85N/m

Extension, e = 0.75m

Mass = 25kg

To find the potential energy of a spring

Potential energy of a spring is given by the formula;

P.E = ½ke²

Substituting into the equation, we have

P.E = ½*85*0.75²

P.E = 42.5 * 0.5625

P.E = 23.91 ≈ 24 Joules

P.E = 24 Joules

8 0
3 years ago
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c
Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0
3 years ago
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