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A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 120 radians/s.What is the speed of ball 1?

Answers

Answer 1

Answer:

The linear speed of the ball for the circular motion is determined as 12 m/s.

The given parameters;

mass of each ball, m = 450 g = 0.45 kg
length of the rod, L = 0.2 m
radius of the rod, r = 0.1 m
angular speed of the ball, ω = 120 rad/s

The linear speed of the ball is calculated as follows;

v = ωr

where;

ω is the angular speed of the ball
r is the radius of circular motion of the ball

The linear speed of the ball is calculated as follows;

v = ωr

v = 120 x 0.1

v = 12 m/s

Thus, the linear speed of the ball for the circular motion is determined as 12 m/s.

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Answer 2

Answer:

Answer:

The speed of ball is 12 $(m)/(s)$

Explanation:

Given:

Mass of ball $m = 0.45$ kg

Radius of rotation $r = 0.1$ m

Angular speed $\omega = 120 (rad)/(s)$

Here barbell spins around a pivot at its center and barbell consists of two small balls,

From the formula of speed in terms of angular speed,

$v = r \omega$

Where $v =$ speed of ball

$v = 120 * 0.1$

$v = 12 (m)/(s)$

Therefore, the speed of ball is 12 $(m)/(s)$

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The animal that is hunted and consumed is considered the

Answers

Predator? like they hunt their prey

Answer:

prey

Explanation:

A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°C. What will be the volume of the air in the bag if you take it with you, still sealed, to the mountains where the temperature is 7.00°C and atmospheric pressure is 70.0 kPa

Answers

The volume of the air in the bag of potato chips to the mountains which is still sealed, 2.766 liters.

What is the gas law?

The gas law is used to show the relationship between the pressure and the temperature of the gases. It can be given as,

$PV=nrT$

Here, (n) and (r) are the constant. Therefore,

$(PV)/(T)=\rm Constant$

For the initial and final values, the gas law can be given as,

$(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$

Here, (subscript 1,and 2) is used for the initial and final amount of pressure and temperature.

The initial values of the bag of potato chips as volume of 2.00 L, pressure of 1.00 ATM and a temperature of 20.0°C. It is known that the value of 1 ATM is equal to the 101.325 kPa.

The final temperature of the pack is 7.00°C and atmospheric pressure is 70.0 kPa

Put the values in the above formula as,

$(101.325*2)/(293)=(70* V_2)/(280)\nV_2=2.766\rm \; liters$

Hence, the volume of the air in the bag of potato chips to the mountains which is still sealed, 2.766 liters.

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Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_(1) = 2.00\ L$

Pressure $P_(1)= 1.00\ atm$

Pressure $P_(2)= 70.0\ kPa$

Temperature $T_(1)= 20.0°C = 293\ K$

Temperature $T_(2)= 7.00°C = 280\ K$

We need to calculate the volume at mountains

Using gas law

$(PV)/(T)=\ Constant$

For both temperature,

$(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))$

Put the value into the formula

$(101.325*2)/(293)=(70* V_(2))/(280)$

$V_(2)=(101.325*2*280)/(293*70)$

$V_(2)=2.766\ litre$

Hence, The volume at mountains is 2.766 L.

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

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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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 current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in the +z direction exists throughout the plane of the loop. The current has the magnitude of I and it is deirected counterclockwise when observing from positive z axis.What is the magnetic force exerted by the external field on the loop?Express your answer in terms of some or all of the variables I, R, and B

Answers

Final answer:

The net magnetic force exerted by the external magnetic field on a current-carrying wire formed into a loop in a uniform magnetic field is absolutely zero since the individual forces on each section of the loop cancel each other out.

Explanation:

The force exerted by a magnetic field on a current carrying wire is given by Lorentz force law, which says that the force is equal to the cross product of the current and the magnetic field. However, in this case, where the wire is formed into a loop with current flowing in a counter-clockwise direction in presence of an external magnetic field, the individual forces on each infinitesimal section of the loop cancel each other out. Therefore, the net magnetic force exerted by the external field on the entire loop is zero.

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Final answer:

The magnetic force exerted on a current-carrying wire loop by an external magnetic field can be calculated using the equation F = I * R * B.

Explanation:

The magnetic force exerted by the external field on the current-carrying wire loop can be determined using the equation F = I * R * B. The magnetic force is equal to the product of the current, radius, and magnetic field strength. The direction of the magnetic force can be determined using the right-hand rule, where the thumb represents the direction of the current, the fingers represent the magnetic field, and the palm represents the direction of the force.

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According to the World Flying Disk Federation, the world distance record for a flying disk throw in the men’s 85-years-and-older category is held by Jack Roddick of Pennsylvania, who on July 13, 2007, at the age of 86, threw a flying disk for a distance of 54.0 m. If the flying disk was thrown horizontally with a speed of 13.0 m/s, how long did the flying disk remain aloft? (Jack Roddick was also a physics teacher! Read more about him at

Answers

Answer:

t = 4.15 seconds

Explanation:

It is given that,

Distance traveled by a flying disk, d = 54 m

The speed at which it was thrown, v = 13 m/s

We need to find the time for which the flying disk remain aloft. Let the distance is d. We know that, speed is equal to the distance covered divided by time. So,

$t=(d)/(v)\n\nt=(54\ m)/(13\ m/s)\n\nt=4.15\ s$

Hence, for 4.15 seconds the flying disk remain aloft.

OFFERING 60 POINTS IF YOU CAN SHOW THE WORK!!!!A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a second hill. As the coaster moves from its initial height to its lowest position, 1700J of energy is transformed to thermal energy by friction.

Answers

Answer; 10.6 i think

Explanation:

(a) At the top of the hill, the coaster has total energy (potential and kinetic)

E = (1000 kg) g (10 m) + 1/2 (1000 kg) (6 m/s)² = 116,000 J

As it reaches its lowest position, its potential energy is converted to kinetic energy, and some is lost to friction, making its speed v such that

1/2 (1000 kg) v ² = 116,000 J - 1700 J = 114,300 J

===> v ≈ 15.2 m/s

If no energy is lost to friction as the coaster makes its way up the second hill, all of its kinetic energy would be converted to potential energy at the maximum possible height H.

1/2 (1000 kg) (15.2 m/s)² = (1000 kg) gH

===> H ≈ 11.7 m

(b) At the top of the second hill with minimum height h, and with maximum speed 4.6 m/s, the coaster has energy

E = P + K = (1000 kg) gh + 1/2 (1000 kg) (4.6 m/s)²

Assuming friction isn't a factor again, the energy here should match the energy at the lowest point in part (a), 114,300 J.

(1000 kg) g h + 1/2 (1000 kg) (4.6 m/s)² = 114,300 J

===> h ≈ 10.6 m

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