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Paraphin [41]
3 years ago
15

A spring is compressed 0.035 m inside a dart gun. (K=500 N/m). The spring has elastic energy. Calculate it.

Physics
1 answer:
NARA [144]3 years ago
6 0

The elastic potential energy of the spring is 0.31 J

Explanation:

The elastic potential energy of a spring is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the compression/stretching of the spring

For the spring in this problem, we have:

k = 500 N/m (spring constant)

x = 0.035 m (compression)

Substituting, we find the elastic potential energy:

E=\frac{1}{2}(500)(0.035)^2=0.31 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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3 years ago
A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?
Olenka [21]
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
F = k*x

Solving: 
F = k*x
F = 50*0.15
\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
E = \frac{k*x^2}{2}

Solving:(Energy associated with this stretching)
E = \frac{k*x^2}{2}
E =  \frac{50*0.15^2}{2}
E =  \frac{50*0.0225}{2}
E =  \frac{1.125}{2}
\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark

7 0
3 years ago
A car stopped at a red light, not moving?
timofeeve [1]

I think it's a) 1st Newton's law... so sorry if it's wrong...

5 0
3 years ago
As you ride a bicycle on the sidewalk with a speed of
mixer [17]

Answer:

1.2 seconds

Explanation:

distance = ((final speed + initial speed) * time)/2

Here given:

  • distance: 3.8 meter
  • initial speed: 6.4 m/s
  • final speed: 0 m/s

Solving steps:

3.8 = ((0 + 6.4) * time))/2

3.8 = 3.2(time)

time = 3.8/3.2

time = 1.1875 seconds ≈ 1.2 seconds

5 0
1 year ago
A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0
IgorLugansk [536]

Answer:

Magnitude of the force is

F_3 = 2.83

direction of the force is given as

\theta = 45 degree West of South

Explanation:

As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy

So here we have

\vec F_1 + \vec F_2 + \vec F_3 = 0

here we know that first force is of magnitude 2 N towards east

\vec F_1 = 2 \hat i N

second force is also of 2.0 N due North

\vec F_2 = 2 \hat j

now from above equation

2\hat i + 2\hat j + \vec F_3 = 0

\vec F_3 = -2\hat i - 2\hat j

so magnitude of the force is given as

F_3 = \sqrt{2^2 + 2^2}

F_3 = 2.83

direction of the force is given as

\theta = tan^{-1}\frac{F_y}{F_x}

\theta = tan^{-1}\frac{-2}{-2}

\theta = 45 degree West of South

3 0
3 years ago
Read 2 more answers
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