Ask question

Ask question

All categories

4 years ago

5

A 56.0kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.50 turns each

second. The distance from one hand to the other is 1.5m. Biometric measurements indicate that each hand typically makes up about 1.25 % of body weight.
Part A) horizontal force must her wrist exert on her hand F=150N

Express the force in part (a) as a multiple of the weight of her hand?

1 answer:

Arada [10]4 years ago

3 0

Answer:

Net force = 129.4, Force as multiple of weight of her hand = 18.84

Explanation:

Given Data:

Total body weight = 56.0 kg ;

no. of turns = 2.5/second ;

hand to hand distance = 1.5m ;

weight of hand = 1.25% of body weight ;

Solution:

mass of hand = $\frac{1.25}{100}$ *56 = 0.7kg ;

radius = d/2 = 1.5/2 = 0.75m ;

Now we need to find velocity, as we know that velocity can be calculated by dividing distance by time

v = d/t = $\frac{2.5*2*3.14*0.75}{1}$ = 11.775 m/s or 12 m/s;

a.

The formula to calculate force is given as

F = mv²/r = (0.7*11.775²)/0.75 = 129.4 N

b.

To calculate force as multiple of weight on her hand, we need to calculate the gravitational force W on her hand first.

W = gm = 9.81 * 0.7 = 6.867 N

Now the wieght on her hand can be represented by

$\frac{Force_{net} }{weight of hand}$ = 129.4 / 6.867 = 18.84

You might be interested in

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

3 years ago

For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d

Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material ( $\sigma$ ) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

Yield Stress = $\frac{Maximum load}{Area of the bar}$

⇒ $\sigma$ = $\frac{P_{max} }{A}$ ---------------- (1)

⇒ Area of the bar (A) = $\frac{\pi}{4}$ × $D^{2}$

⇒ A = $\frac{\pi}{4}$ × $45^{2}$

⇒ A = 1589.625 $mm^{2}$

Put all the values in equation (1) we get

⇒ $P_{max}$ = 200 × 1589.625

⇒ $P_{max}$ = 317925 N

In this bar the $P_{max}$ is equal to the weight of the bar.

⇒ $P_{max}$ = $M_{max}$ × g

Where $M_{max}$ is the maximum mass the bar can support.

⇒ $M_{max}$ = $\frac{P_{max} }{g}$

Put all the values in the above formula we get

⇒ $M_{max}$ = $\frac{317925}{9.81}$

⇒ $M_{max}$ = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

3 0

3 years ago

Which instrument is used to measure liquid volume?

Alecsey [184]

Volumetric cylinders and volumetric flasks

3 0

3 years ago

Read 2 more answers

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

3 years ago

A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m + t/(0.20 s))], where x is in m and t in s.

Len [333]

Corrected and Formatted Question:

A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))], where x is in m and t in s.

(a) In what direction is this wave traveling?

(b) What are the wave speed, frequency, and wavelength?

(c) At t = 0.50 , what is the displacement of the string at x = 0.20 m?

Answer:

The wave is travelling in the negative x direction

The wave speed = 12.0m/s

The frequency = 5Hz

The wavelength = 2.4m

The displacement at t = 0.50s and x = 0.20m is -0.029m

Explanation:

The general wave equation is given by;

y(x, t) = y cos (2 $\pi$ (x/λ) - 2 $\pi$ ft) --------------------------------(i)

Where;

y(x, t) is the displacement of the wave at position x and a given time t

y = amplitude of the wave

f = frequency of the wave

λ = wavelength of the wave

Given;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))] ------------------(ii)

Which can be re-written as;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m)) + 2π(t/(0.20 s))] -------------(iii)

Comparing equations (i) and (iii) we have that;

=> 2π(x/(2.4 m) = 2π(x/λ)

=> λ = 2.4m

Therefore the wavelength of the wave is 2.4m

Also, still comparing the two equations;

=> 2π(t/(0.20 s) = 2πft

=> f = 1 / 0.20

=> f = 5Hz

Therefore the frequency of the wave is 5Hz

To get the wave speed (v), it is given by;

v = f x λ

Where f = 5Hz and λ = 2.4m

=> v = 5 x 2.4

=> v = 12.0m/s

Therefore, the speed of the wave is 12.0m/s

At t = 0.50s and x = 0.20m;

The displacement, y(x,t) of the string wave is given by

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]

<em>Convert the amplitude of 3.0cm to m</em>

=> 3.0cm = 0.03m

<em>Substitute this back into the equation</em>

=> y(x, t) = (0.03m) × cos[2π(x/(2.4 m) + t/(0.20 s))]

<em>Substitute the values of t and x into the equation above;</em>

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m) + 0.50/(0.20 s))]

<em>Carefully solve the equation</em>

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m)) + 2π(0.50/(0.20 s))]

=> y(x, t) = (0.03m) × cos[0.08π + 5π]

=> y(x, t) = (0.03m) × cos[5.08π]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × -0.9684

=> y(x, t) = 0.029m

Therefore the displacement at those points is -0.029m

Also, the sign of the displacement shows that the direction of the wave is in the negative x direction.

8 0

3 years ago