Please help me with this and if I have good answer u will be marked as brilliant !!!!

A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A

Ghella [55]

The total work W done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

W = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

W = ∆K = K

where K is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

W = 1/2 mv ²

where m is the mass of the object and v is the speed you want to find. Solving for v, you get

v = √(2W/m) ≈ 0.46 m/s

8 0

3 years ago

Who was the creator of sports

Elden [556K]

The first Olympic Games were in Greece

8 0

3 years ago

Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro

Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

7 0

3 years ago

Heather and Matthew walk with an average velocity of 0.98 m/s eastward. If it takes them 34 min to walk to the store, what is th

TEA [102]

Answer:

D= 1999.2 m

Explanation:

Given that

Average velocity ,v= 0.98 m/s

time ,t= 34 min

We know that

1 min = 60 s

That is why

t= 34 x 60 =2040 s

We know that

Displacement = Average velocity x time

D= v t

Now by putting the values in the above equation

D= 0.98 x 2040 m

D= 1999.2 m (eastward)

The direction of the displacement will be towards eastward.

That is why the displacement will be 1999.2 m or we can say that 1.9992 km.

3 0

3 years ago

Show all work.

lys-0071 [83]

The new gravitation force at the new location is 40 N

Explanation:

The weight of the astronaut is given by the equation

$F=mg$ (1)

where

m is the mass of the astronaut

g is the acceleration of gravity

The acceleration of gravity at a certain distance $r$ from the centre of the Earth is given by

$g=\frac{GM}{r^2}$

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

$F=\frac{GMm}{r^2}$

When the astronaut is on the Earth's surface, $r=R$ (where R is the Earth's radius), so his weight is

$F=\frac{GMm}{R^2}=640 N$

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

$r'=3R+R=4R$

Therefore, the new weight is

$F'=\frac{GMm}{(4R)^2}=\frac{1}{16}\frac{GMm}{R^2}=\frac{F}{16}$

Which means that his weight has decreased by a factor 16: therefore, the new weight is

$F'=\frac{640}{16}=40 N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

3 0

3 years ago