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3 years ago

In Hooke's law, Fspring=kΔx , what does the Fspring stand for?

Physics

2 answers:

Mrrafil [7]3 years ago

6 0

Answer:

The amount of force acting on the spring.

Explanation:

Here, Fspring= kΔx is a equation denoting the amount of force acting on the spring.

where, k = spring constant.

Δx= change in the length of the spring.

so, for every Δx change in spring length, kΔx force acts on the spring.

Agata [3.3K]3 years ago

6 0

Answer:

The amount of force acting on the spring.

Explanation:

Here,

Fspring= kΔx is a equation denoting the amount of force acting on the spring.

where, k = spring constant.

Δx= change in the length of the spring.

so, for every Δx change in spring length, kΔx force acts on the spring.

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- CP

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What are three key aerodynamics principles?

mafiozo [28]

Answer:

Explanation:

There are three basic forces in aerodynamics: acceleration, which moves an airplane forward; drag, which holds it back; and height, which keeps it airborne. Lift is generally explained by three theories: Bernoulli's principle, the Coanda effect, and Newton's third law of motion.

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3 years ago

An occupant of a car has a chance of surviving a crash if the deceleration during the crash is not more then 30 g. Calculate the

sweet [91]

Answer:

20,850 N

Explanation:

We can solve the problem by using second Newton's Law:

$F=ma$

where

F is the force

m is the mass

a is the acceleration

In this problem, we have:

m = 70 kg is the mass

$a=-30 g=-30 (9.8 m/s^2)=-294 m/s^2$ is the acceleration (which is negative, because it is a deceleration)

So, we can use the equation above to find the force:

$F=(70 kg)(-294 m/s^2)=-20580 N$

and the negative sign simply means that the force is in the opposite direction to the motion.

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2 years ago

With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by

lara31 [8.8K]

According to the net force, the acceleration of the book is 16.47 m/s².

We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as

∑F = m . a

where F is force, m is mass and a is acceleration

From the question above, we know that

m = 3 kg

g = 9.8 m/s²

F1 = 20 N

Find the net force

∑F = F1 + W

∑F = 20 + m . g

∑F = 20 + 3 . 9.8

∑F = 20 + 29.4

∑F = 49.4 N

Find the acceleration

∑F = m . a

49.4 = 3 . a

a = 16.47 m/s²

Find more on force at: brainly.com/question/25239010

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10 months ago

How many centimeters are in 24 meters

aleksandr82 [10.1K]

The answer is 2400 centimeters

5 0

3 years ago

Read 2 more answers

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

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2 years ago

Read 2 more answers