Define the strip attached to the magnet?

A little girl is swinging on her tire swing. At the lowest point of her swing, the tension in the rope is a) Equal to the gravit

Elodia [21]

A because the girl in that instant is not moving up or down so
( up forces)=(down forces )
The up forces is the tension of the rope and down forces us mg -the gravitational force on the girl by the earth

6 0

3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

3 years ago

At its natural resting length, a muscle is close to its optimallength for producing force. As the muscle contracts, the maximumf

lesantik [10]

Answer:

Figure E is the correct representation of the first part of the motion. When in a hanging position from the chin-up bar, the bicep muscles are stretched beyond their normal length already. So at this point they are at the peak of their capacity and you are at rest (this corresponds to the velocity v = 0 at t = 0). On contracting the bicep muscles and pulling your whole body up, you begin to gain speed and v increases. This increase in velocity is exponential. Soon the bicep muscles contract up to 80% their normal length reducing the force they can produce to keep you rising up to zero. The velocity change happens because the body is accelerating and the muscles can still supply a net force to lift you up. The acceleration is present because of this net force. The moment this force reduces to zero, the acceleration too reduces to zero. (From Newton's second law of motion). This reduction in acceleration is responsible for the reduction of the curvature of the v curve in figure E above. The point where the velocity becomes horizontal corresponds to the point where the muscles reach their maximum contraction unit and can supply no more net force and as a result no acceleration. This further results inba constant velocity which is the flat nature of the curve seen in diagram E.

Thank you for reading.

Explanation:

5 0

3 years ago

A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology a

Zigmanuir [339]

Answer:

5.95 m

Explanation:

Given that the biggest loop is 40.0 m high. Suppose the speed at the top is 10.8 m/s and the corresponding centripetal acceleration is 2g

For the car to stick to the loop without falling down, at the top of the ride, the centripetal force must be equal to the weight of the car. That is,

(MV^2) / r = mg

V^2/ r = centripetal acceleration which is equal to 2g

2 × 9.8 = 10.8^2 / r

r = 116.64 /19.6

r = 5.95 m

3 0

3 years ago

A diver running 2.5 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. how hi

Elodia [21]

<u>Answer:</u>

a) Height of cliff = 44.145 meter

b) 7.5 meter far from its base the diver hit the water.

<u>Explanation:</u>

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Diver's vertical motion:

Initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate displacement when time is 3 seconds.

$s= ut+\frac{1}{2} at^2\\ \\ =0*3+\frac{1}{2} *9.81*3^2\\ \\ =44.145meter$

So height of cliff = 44.145 meter.

b) Diver's horizontal motion:

Initial velocity = 2.5 m/s, acceleration = 0 $m/s^2$ , we need to calculate displacement when time is 3 seconds.

$s= 2.5*3+\frac{1}{2}*0*3^2\\ \\ =7.5meter$

So 7.5 meter far from its base the diver hit the water.

5 0

3 years ago

Define the strip attached to the magnet?​

Define the strip attached to the magnet?