On average, both arms and hands together account for 13 % of a person's mass, while the head is 7.0% and the trunk and legs account for 80 % . We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 61.0-kg skater is 1.80 m tall, has arms that are each 70.0 cm long (including the hands), and a trunk that can be modeled as being 35.0 cm in diameter. The skater is initially spinning at 70.0 rpm with his arms outstretched.Required:
What will his angular velocity be (in rpm) when he pulls in his arms until they are at his sides parallel to his trunk?

Answers

Answer 1

Answer:

Final answer:

To find the final angular velocity when the skater pulls in his arms, we use the conservation of angular momentum.

Explanation:

To find the final angular velocity when the skater pulls in his arms, we can make use of the conservation of angular momentum. Initially, the skater's arms are outstretched, and the moment of inertia can be calculated using the parallel axis theorem. After the skater pulls in his arms, we can calculate the new moment of inertia using the same theorem. Equating the initial and final angular momentum values, we can solve for the final angular velocity.

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Answer 2

Answer:

Final answer:

The problem involves the concept of conservation of angular momentum. The skater's spinning speed will increase when they pull their arms in. For a precise value of the final velocity, a complex calculation taking into account body mass distribution is needed.

Explanation:

This question involves the principle of conservation of angular momentum, which states that the angular momentum of an object remains constant as long as no external torques act on it. The total initial angular momentum of the skater spinning with outstretched arms is equal to his final angular momentum when he pulls his arms in.

Calculating the skater's initial and final angular momentum, you can then solve for his final velocity.

However, note that the calculation needs to take into account the skater's mass distribution. Specifically, we need to consider the percentage distributions for the arms/hands (13%), head (7%) and trunk/legs (80%), and integrate these over the skater's body.

This can result in a significantly complex calculation if done accurately, involving calculus level mathematics. However, using the qualitative knowledge that the skater's spinning speed will increase when they pull their arms in, it's reasonable to estimate, considering the mass distribution, the final velocity will be somewhere near 2 to 3 times the original rpm. But for an exact value, a detailed and complex calculation is needed.

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A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the instant when 28 cm of string has unwound off the disk, what is the torque exerted about the center of the disk?

Answers

The torque exerted about the center of the disk is 0.2845Nm (rounded to 4 decimal places).

What do you mean by torque?

Torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. It represents the capability of a force to produce change in the rotational motion of the body.

Moreover, torque is a twisting or turning force that tends to cause rotation around an axis, which might be a center of mass or a fixed point. Torque can also be thought of as the ability of something that is rotating, such as a gear or a shaft, to overcome turning resistance.

Therefore, torque is defined as Γ=r×F=rFsin(θ). In other words, torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector, 'a' being the angle between r and F.

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Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = $(28)/(10)$ × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?

Answers

Answer:

(a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

Explanation:

Given that,

Initial velocity = 50 km/s

Electric field = 50 N/C

Time = 1.5 ns

(a). We need to calculate the speed of the electron 1.5 n s after entering this region

Using newton's second law

$F = ma$ .....(I)

Using formula of electric force

$F = qE$ .....(II)

from equation (I) and (II)

$-qE= ma$

$a = (-qE)/(m)$

(a). We need to calculate the speed of the electron

Using equation of motion

$v = u+at$

Put the value of a in the equation of motion

$v = 50*10^(3)-(1.6*10^(-19)*50)/(9.1*10^(-31))*1.5*10^(-9)$

$v=36813.18\ m/s$

$v =3.68*10^(4)\ m/s$

(b). We need to calculate the distance traveled by the electron

Using formula of distance

$s = ut+(1)/(2)at^2$

Put the value in the equation

$s = 3.68*10^(4)*1.5*10^(-9)-(1)/(2)*(1.6*10^(-19)*50)/(9.1*10^(-31))*(1.5*10^(-9))^2$

$s=0.0000453\ m$

$s=4.53*10^(-5)\ m$

Hence, (a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

PLEASE HELP DUE BEFORE 11:30 TODAY!!!!Which of the following quantities is NOT a vector quantity?
A. 926 m to the north
B. 5.2 m/s to the west
C. 46 m down
D. 12.3 m/s faster

Answers

Answer:

D is not the a vector quantities

What is a light year

Answers

Answer:

A light-year is the distance light travels in one year.

Answer:

Explanation:

a unit of astronomical distance equivalent to the distance that light travels in one year, which is 9.4607 × 1012 km (nearly 6 million million miles).

A thin film soap bubble (n=1.35) is floating in air. If the thickness of the bubble wall is 300nm, which of the following wavelengths of visible light is strongly reflected?