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jasenka [17]
3 years ago
5

Work of 2 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (i

n newtons) that holds the spring stretched at the same distance (11 cm)
Physics
1 answer:
Sidana [21]3 years ago
5 0
The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.

If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is

W = k[{x₂-x₁)² - (x₁-xn)²],

where 
xn is the natural length
x₁ is the stretched length 
x₂ is also the stretched length when stretched even further than x₁

In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.

Then, substituting the values,

2 J = k (0.11² -0²)
k = 165.29 N/m

Finally, we use the value of k to the Hooke's Law to determine the Force.

F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
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A pillow with mass of 0.3 kg sits on a bed with a coefficient of static friction of 0.6. What is the maximum force of static fri
lys-0071 [83]
The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body. 
 
                        P = W = mass x acceleration due to gravity
    
                    P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N

Solving for the static friction force (F), 
                                              F = P x c 
 
                                      F = (2.94 N) x 0.6 = 1.794 N

Therefore, the maximum force of static friction is 1.794 N. 



5 0
3 years ago
Read 2 more answers
A 50.0 kg woman climbs a flight of stairs 6.00 m high in 15.0 s. How much power does she use.
WITCHER [35]

Her weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

Work = (weight) · (height) = (50kg) · (9.8 m/s²) · (6 m)

Power = (work) / (time) = (50kg) · (9.8 m/s²) · (6 m) / (15 s)

Power = (50 · 9.8 · 6 / 15) · (kg · m² / s³)

Power = 196 (kg · m / s²) · (m) / s

Power = 196 Newton-meter/second

<em>Power = 196 watts</em>

6 0
3 years ago
A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa
kotykmax [81]
Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is v_i=180~m/s, the final velocity is v_i=0~m/s, and the total time of the motion is \Delta t=0.02~s, so the acceleration is given by
a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2 
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m
So, the bullet penetrates the sandbag 1.8 meters.
5 0
3 years ago
What does itmean to have an acceleration of 8 m/s2
Masteriza [31]

Answer:

The answer is option A.

You speed up 8 m/s every second

Hope this helps you

8 0
2 years ago
Read 2 more answers
in the diagram, q2 is +34.4*10^-6 C, and q3 is -72.8*10^-6 C. The net force on q2 is 225 N to the right. What is q1? Include the
sesenic [268]

Answer:

  q₁ = -6.54 10⁻⁵ C

Explanation:

Force is a vector quantity, but since all charges are on the x-axis, we can work in one dimension, let's apply Newton's second law

                F = F₁₂ + F₂₃

the electric force is given by Coulomb's law

                F = k q₁q₂ / r₁₂²

let's write the expression for each force

                F₂₃ = k q₂ q₃ / r₂₃²

                F₂₃ = 9 10⁹ 34.4 10⁻⁶ 72.8 10⁻⁶ / 0.1²

                F₂₃ = 2.25 10³ N

               

                F₁₂ = k q₁q₂ / r₁₂²

                F₁₂ = 9 10⁹ q₁ 34.4 10⁻⁶ / 0.1²

                F₁₂ = q₁   3,096 10⁷ N

we substitute in the first equation

                225 = q₁  3,096 10⁷ +2.25 10³

                q₁ = (225 - 2.25 10³) / 3,096 10⁷

                q₁ = -6.54 10⁻⁵ C

4 0
3 years ago
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