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iVinArrow [24]
3 years ago
14

El muelle de un dinamómetro se alarga 6cm cuando aplicamos sobre él una fuerza de 10 N. Calcula el alargamiento del muelle al ap

licar una fuerza de 20 N.
Physics
1 answer:
Crank3 years ago
6 0

Answer:

x = 12cm

Explanation:

First, with the values given for the force and the displacement, you calculate the spring-constant of the dynamometer by using the following equation (primero, con los valores dados de la fuerza y el desplazamiento, calculas la constante del resorte aplicando la siguiente formula):

F=kx

F: force (fuerza) = 10N

k: spring constant (constante del resorte)

x: displacement of the spring (desplazamiento del resorte) = 6cm

by using these values you calculate k (usando estos valores calculas k):

k=\frac{F}{x}=\frac{10N}{6cm}=1.66\frac{N}{cm}

With this value you calculate the displacement for a force of 20N (con este valor de k calculas el desplazamiento del resorte para una fuerza de 20N):

x=\frac{F}{k}=\frac{20}{1.66N/cm}=12cm

hence, the displacement is 12cm

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Temperature which is solid becomes a liquid
Brums [2.3K]

Answer:

The answer would be melting point.

Explanation:

Hope this helps. Can you please mark me brainliest

5 0
3 years ago
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A plastic tube allows a flow of 15.9 cm3 /s of water through it. how long will it take to fill a 237 cm3 bottle with water? answ
xxMikexx [17]
You are given a fixed rate of 15.9 cm³/s. You are also given with the amount of volume in 237 cm³. Through the approach of dimensional analysis, you can manipulate through operations such that the end result of the units must be in seconds. The solution is as follows:

237 cm³ * (1 s/15.9 cm³) = 14.9 seconds
7 0
3 years ago
What kind of charge is produced when a silk shirt is rubbed with human body<br>​
jeka94

Answer:

Frictional charges

Explanation:

6 0
3 years ago
ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself o
shtirl [24]

Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver

Answer:

-6461.54 N

Explanation:

From Newton's Fundamental equation,

F = m(v-u)/t.................... Equation 1

Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.

Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s

Substitute into equation 1

F = 0.75(-60-52)/0.013

F = 0.75(-112)/0.013

F = -84/0.013

F = -6461.54 N

Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.

Hence the magnitude of the average force of the wall = -6461.54 N

4 0
3 years ago
a spring is stretched 20 cm by a 30.0 n force. determine the work done in stretching the spring from 0 to 40 cm?
Alekssandra [29.7K]

The work done when a spring is stretched from 0 to 40cm is 4J.

What is work done?

Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.

The work done on the spring to stretch to 40cm is,

F = kx

where F is force, k is force constant.

k = F / x = 10 N / 20 * 10^-2 m = 50 N/m

W = 0.5 * k * (x)^2

where W = work done, k = force constant.

W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.

Therefore, the work done on the spring when it is stretched to 40cm is 4J.

To learn more about work done click on the given link brainly.com/question/25573309

#SPJ4

3 0
1 year ago
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