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Eva8 [605]
3 years ago
13

An object with mass 0.900kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00m to the rig

ht (the x - direction) to stretch the spring, and released. What is the speed of the object when it is 0.50m to the right of the x
Physics
1 answer:
Neko [114]3 years ago
7 0

Answer:

7.85 m/s

Explanation:

We are given that

Mass of object=m=0.900 kg

F(x)=\alpha x-\beta x^2

\alpha=60 N/m

\beta=18N/m^2

F(x)=-60x-18x^2

U=0 when x=0

Potential energy=-\int F(x)dx

Substitute the values

U(x)=-\int (-60x-18x^2)dx

U(x)=60(\frac{x^2}{2})+18(\frac{x^3}{3})+C

Using the formula

\int x^n dx=\frac{x^{n+1}}{n+1}+C

Substitute x=0

U(0)=C\implies C=0

U(x)=30x^2+6x^3

x_1=0.5,x_2=1

v_2=0

Using law of conservation energy

\frac{1}{2}mv^2_1+U(x_1)=\frac{1}{2}mv^2_2+U(x_2)

Substitute the values

\frac{1}{2}(0.9)v^2_1+30(0.5)^2+6(0.5)^3=0+30(1)^2+6(1)^3

\frac{1}{2}(0.9)v^2_1+8.25=36

\frac{1}{2}(0.9)v^2_1=36-8.25=27.75

v^2_1=\frac{27.75\times 2}{0.9}

v_1=\sqrt{\frac{27.75\times 2}{0.9}}

v_1=7.85 m/s

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Rama09 [41]

Answer:

12 straps

Step-By-Step Explanation:

84 - 32 - 29 - 15 =

52 - 29 - 15 =

23 - 15 = 12

btw this is supposed to be in maths category

<em><u>If this helped, please consider picking this answer as the Brainliest Answer. Thank you!</u></em>

3 0
3 years ago
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 998
wlad13 [49]

Answer:

<em>1988.05 rad/s^2</em>

<em></em>

Explanation:

The angular speed of the optical disk ω = 998.0 rad/s

the time taken to come to rest t = 0.502 s

The magnitude of the average angular acceleration ∝ = ω/t

∝ = 998.0/0.502 = <em>1988.05 rad/s^2</em>

7 0
3 years ago
It takes 6 seconds for a stone to fall to the bottom of a mine shaft. how deep is the shaft
Margaret [11]
You need the kinematic equation for distance as a function of acceleration:

d = [v(initial) *t] + 0.5a*t^{2}

where a = gravitational acceleration 9.8m/s^{2}
v(initial) = starting velocity
t = time of fall

if the stone started at rest then v(initial) = 0 making the equation simply 0.5a * t^{2}

Let me know if you still need further help :)
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Draw the ray diagram of Radius of concave mirror=7cm
seropon [69]

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Nature and size of image:

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8 0
3 years ago
what is the magnitude of the compression forces (assumed to be horizontal) acting on both sides of the center board that is sand
Len [333]

F = normal force by each board on each side

W = weight of the board in between acting in down direction = 95.5 N

f = frictional force in upward direction by each board

\mu = coefficient of friction = 0.663

Using equilibrium of force in Upward direction

f + f = W

f = W/2

f = 95.5/2 = 47.75 N

frictional force is given as

f = \mu F

47.75 = (0.663) F

F = 72.02 N

4 0
3 years ago
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