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zlopas [31]
3 years ago
13

An unstretched spring has a length of 10. centimeters. When the spring is stretched by a force of 16 newtons, its length is incr

eased to 18 centimeters. What is the spring constant of this spring
Physics
1 answer:
Fed [463]3 years ago
5 0

Answer:

The spring constant of this spring  is 200 N/m.

Explanation:

Given:

Original unstretched length of the spring (x₀) = 10 cm =0.10 m [1 cm =0.01 m]

Stretched length of the spring (x₁) = 18 cm = 0.18 cm

Force acting on the spring (F) = 16 N

Spring constant of the spring (k) = ?

First let us find the change in length of the spring or the elongation caused in the spring due to the applied force.

So, Change in length = Final length - Initial length

\Delta x = x_1-x_0=0.18-0.10=0.08\ m

Now, restoring force acting on the spring is directly related to its elongation or compression as:

F=k\Delta x

Rewriting in terms of 'k', we get:

k=\dfrac{F}{\Delta x}

Now, plug in the given values and solve for 'k'. This gives,

k=\frac{16\ N}{0.08\ m}\\\\k=200\ N/m

Therefore, the spring constant of this spring  is 200 N/m.

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water has an index of refraction of 1.33. What is the critical angle for light leaving a pool of water into air? 0 37 90 49
vesna_86 [32]
When light travels from a medium with higher refractive index into a medium with lower refractive index, there is a maximum angle (called critical angle) for which all the light is reflected, so there is no refraction.

The value of the critical angle is given by:
\theta = \arcsin ( \frac{n_2}{n_1} )
when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
\theta = \arcsin ( \frac{1.00}{1.33} )=49^{\circ}
6 0
3 years ago
A basketball player shoots toward a basket 5.3 m away and 3.0 m above the floor. If the ball is released 1.9 m above the floor a
Snezhnost [94]

Answer:

Vi = 8.28 m/s

Explanation:

This problem is related to the projectile motion.

As we know there are two components of motion associated with this, the horizontal component and vertical component.

The horizontal distance covered by the ball is

Vx*t = x

Vx*t = 5.3

Vx = 5.3/t  eq. 1

Also we know that

Vx = Vicos(60)

Vx = Vi*0.5  eq. 2

equate eq. 1 and eq. 2

5.3/t = Vi*0.5

5.3/0.5 = Vi*t

Vi*t = 10.6  eq. 3

The vertical distance is

Vy = y1 + Vyi*t - 0.5gt²

also we know that

Vyi = Visin(60)

Vyi = Vi*0.866

It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance

3 = 1.9 + Vi*0.866*t - 0.5gt²

3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

1.1 = 0.866(Vi*t) - 4.9t²

0.866(Vi*t) = 4.9t² + 1.1

substitute Vi*t = 10.6 in above equation

0.866(10.6) = 4.9t² + 1.1

9.18 = 4.9t² + 1.1

4.9t² = 8.08

t² = 8.08/4.9

t² = 1.648

t = 1.28 sec

Finally, initial speed can be found by substituting the value of t into eq. 3

Vi*t = 10.6

Vi = 10.6/t

Vi = 10.6/1.28

Vi = 8.28 m/s

8 0
3 years ago
Pravat exerts a force of 30 N to lift a bag of groceries 0.5 m. How much work did Pravat do on the bag?
Dahasolnce [82]

Answer:

15 J

Explanation:

Work = Force x Distance

15= 30 x 0.5

Have a blessed day!

7 0
3 years ago
Read 2 more answers
A mass m is tied to an ideal spring with force constant k and rests on a frictionless surface. The mass moves along the x axis.
7nadin3 [17]

Answer:x=\frac{x_m}{\sqrt{2}}

Explanation:

Given

initially mass is stretched to x_m

Let k be the spring Constant of spring

Therefore Total Mechanical Energy is \frac{kx_m^2}{2}

Position at which kinetic Energy is equal to Elastic Potential Energy

K=\frac{mv^2}{2}

U=\frac{kx^2}{2}

it is given

k=U

thus 2U=\frac{kx_m^2}{2}

2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}

2x^2=x_m^2

x=\frac{x_m}{\sqrt{2}}

3 0
3 years ago
A proton of mass 1.67 x 10-27 kg (the charge of a proton is 1.6 x 10-19 C) enters the region between two parallel plates a
Zepler [3.9K]
The answer to the question is 72773
3 0
2 years ago
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