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g100num [7]
3 years ago
10

(1 pt) this problem is an example of critically damped harmonic motion. a hollow steel ball weighing 4 pounds is suspended from

a spring. this stretches the spring 18 feet. the ball is started in motion from the equilibrium position with a downward velocity of 4 feet per second. the air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) . suppose that after t seconds the ball is y feet below its rest position. find y in terms of t. take as the gravitational acceleration 32 feet per second per second. (note that the positive y direction is down in this problem.) y=
Physics
1 answer:
Zepler [3.9K]3 years ago
3 0
<span>Answer: Hooke's Law along with Newtons second Law of motion. x"+(a/m)x'+(k/m)x=F(t)/m x"+(256)x'+(96)x=0 F=-kx F=-4 x=1/3 mg=4 m=4/32 m=1/8 -4=-k* (1/3) k=12 x"+(256)x'+(96)x=0 y=C1e^(4*(1018)^(1/2)-128)*t+C2e^(-4*(...</span>
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A man walks 400 m in the direction 45° north of east. Represent this vector graphically by
Marrrta [24]

Answer:

Explanation:

Coordinate system is one that describe the location of an object in a given plane. It implies the use of axes (coordinates) and points.

Given that the man in the question walks 400 m due north of east. The cardinal points can be used in this case, with the north and east cardinals as the required axis.

scale = \frac{length on drawing}{original length}

         = \frac{10}{400}

         = \frac{1}{40}

scale = 1:40

This is a reduced scale which implies that 1 cm on the drawing is equal to 40 m on the original length.

The man's direction is 45^{o} north of east.

The graphical drawing of the vector is herewith attached to this answer.

5 0
3 years ago
A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/mN/m. At t=0 t=0 the block has velocity -
slega [8]

Answer:

The amplitude of the spring is 32.6 cm.

Explanation:

It is given that,

Mass of the block, m = 2 kg

Force constant of the spring, k = 300 N/m

At t = 0, the velocity of the block, v = -4 m/s

Displacement of the block, x = 0.2 mm = 0.0002 m

We need to find the amplitude of the spring. We know that the velocity in terms of amplitude and the angular velocity is given by :

v=\omega\sqrt{A^2-x^2}

\omega=\sqrt{\dfrac{k}{m}}

\omega=\sqrt{\dfrac{300}{2}}    

\omega=12.24\ rad/s

So, \dfrac{v^2}{\omega^2}+x^2=A^2

\dfrac{(-4)^2}{(12.24)^2}+(0.0002)^2=A^2            

A = 0.326 m

or

A = 32.6 cm

So, the amplitude of the spring is 32.6 cm. Hence, this is the required solution.

4 0
4 years ago
Someone Pls help me it will help. I will give brainlist.
Afina-wow [57]

Answer:

nm i believe pls crown

Explanation:

5 0
3 years ago
The earth's magnetic moment is 8 x 1022 a-m2. If that moment were created by a loop of wire going around the earth (r = 6378 km)
Elena L [17]

Answer:

M = I A      definition of magnetic moment - current * area

A = π R^2 = π * (6.4E6)^2 = 1.3E14 m^2

I = 8E22 A-m^2 / 1.3E14 m^2 = 6.2E8 amperes

I = 620,000,000 amps

4 0
2 years ago
A ball is dropped from rest and falls to the floor. The initial gravitational potential energy of the ball-Earth-floor system is
Cloud [144]

Answer:

The mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.

Explanation:

It is given that,

Initial gravitational potential energy of the ball-Earth-floor system is 10 J.

The ball then bounces back up to a height where the gravitational potential energy is 7 J.

Let U is the mechanical energy of the ball-Earth-floor system the instant the ball left the floor. Due to the conservation of energy, the mechanical energy is equal to difference between initial gravitational potential energy and the after bouncing back up to a height.

Initial mechanical energy is 10 + 0 = 10 J

Mechanical energy just before the collision is 0 + 10 = 10 J

Final mechanical energy, 7 + 0 = 7 J

Hence, the mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.

4 0
3 years ago
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