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Sav [38]
2 years ago
11

A spring with a spring constant of 350 N/m pulls a door closed. How much work is done as the spring pulls the door at a constant

velocity from an 85.0-cam stretch to a 5.0-cm stretch?
Physics
1 answer:
german2 years ago
3 0

The work done to stretch the spring will be 112 J.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =?

K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

The work is done to stretch the spring is;

\rm W= \frac{1}{2} kx^2 \\\\ W=\frac{1}{2} \times 350 \times (0.850-0.050)^2 \\\\ W=0.5 \times 350 \times (0.80)^2 \\\\W=112 \ J

To learn more about the spring force refer to the link;

brainly.com/question/4291098

#SPJ1

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A cup of tea at room temperature of 24°C is heated until it has twice the internal energy. Calculate the final temperature of th
elena-s [515]

The final temperature of the tea cup is 100°C.

<h3>What is internal energy?</h3>

The Internal energy is the energy of a substance due to to the constant random motion of its particles.

The symbol for internal energy of a substance is U and it is measured in Joules.

ΔU = q + W

  • q is the heat, q = mcΔT
  • W is the mechanical work.

In conclusion,  the final temperature of the tea cup at room temperature of 24 °C which is heated until it has twice the internal energy is 100°C.

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3 0
2 years ago
Suppose the U.S. national debt is about $15 trillion. If payments were made at the rate of $1,500 per second, how many years wou
Delvig [45]

Answer:

This question has already been answered.

Explanation:

brainly.com/question/13542582

3 0
3 years ago
A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0
3 years ago
You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 35 m/s. The ball has __________________
Elden [556K]

Answer:

The ball has kinetic energy

the kinetic energy is 945 J

Explanation:

4 0
3 years ago
Read 2 more answers
3. a) Your body is made up of several simple machines that help you move. Identify three
Alja [10]

Answer:

Explained below

Explanation:

1) The human arm: This is a type of simple machine called "Lever". In this type of machine, the elbow acts as the fulcrum, the palm serves as the load because that's where we place the load we want to carry. While the inner part of the arm which is the inner part of the elbow represents the effort because that is the joint we mover when making use of our arms.

2) Pulleys: An example of this in the human body is the knee cap where the direction of an applied force is changed. Thus means as it is in motion, it alters the direction for which the quadriceps tendon pulls on the tibia.

3) wheel and axle: An example of this in the human body is the lateral rotation of the shoulder joint medial. The humerus which is the bone between the shoulder and elbow will act as the axle while the rotator will be the will because when it is rotated a little bit, the humerus will move along with it.

8 0
3 years ago
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