You hit a hockey puck and it slides across the ice at nearly a constant speed.Is a force keeping it in motion?Explain.

Answers

Answer 1

Answer:

At constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.

According to Newton's second law of motion, the force applied to a an object is directly proportional to the product of mass and acceleration of the object.

F = ma

Acceleration is the change in the velocity of an object per change in time of motion.

At constant velocity, the acceleration of an object iszero.

When acceleration of an object is zero, the force on the object is zero.
A constant speed (magnitude only) and change in the direction of the object, implies a change in velocity of the object.
at changing velocity, the acceleration on an object is positive, and hence net force acts on the object.

Thus, we can conclude that at constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.

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Answer 2

Answer:

Answer:

Explanation:

When the puck is sliding on the ice, there is no force being exerted on the puck to keep it moving forward. Instead, inertia keeps the puck moving forward. Friction between the puck and the ice gradually slows the puck down. You hit a hockey puck and it slides across the ice at nearly a constant speed

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Susan’s 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul’s speed after being pulled 3.0 m.

Answers

Final answer:

In this problem, we have calculated the work done by Susan pulling her baby brother on a mat and the work done against friction. The net work done, which is the work done by Susan's pulling minus the work done against friction, is transformed into kinetic energy, giving us the baby's speed after being pulled 3m, which is approximately 1.95 m/s.

Explanation:

To answer this question, we first need to calculate the work done by Susan when she pulls the mat over the distance of 3.0 meters. The angle at which the rope is pulled does make a difference in this calculation. The force that is actually contributing to the work is the horizontal component of the tension, which can be determined by the equation Fh = F cos θ which equals 30N * cos30 = 25.98N.

The work done, W, is equal to this force multiplied by the displacement, so W = Fd = 25.98N * 3m = 77.94 Joules.

Next, we need to calculate the work done against friction. The force of friction is calculated as Ff = µN. Here N is the normal force, which is equal to the weight of the baby, so N = mg = 10kg * 9.8m/s² = 98N. The force of friction then is Ff = µN = 0.20 * 98N = 19.6N. The work done against friction is Wf = Ff * d = 19.6N * 3m = 58.8 Joules.

The net work done on the baby is the work done by Susan minus the work done against friction, so Wnet = W - Wf = 77.94J - 58.8J = 19.14 Joules. This net work is equal to the change in kinetic energy of the baby, ∆K, since Kinitial = 0 (Paul starts at rest), the work done is all transformed into final kinetic energy. So ∆K = 19.14J.

The kinetic energy of an object is given by the equation K = 1/2 mv², so we have 19.14J = 1/2 * 10kg * v². Solving for v gives us roughly v = 1.95 m/s. Therefore, the speed of the baby after being pulled 3 meters is approximately 1.95 m/s.

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Final answer:

To determine Paul's speed, we must calculate the net work done on him using the work-energy theorem. This includes the work done by Susan and the work done against friction. Paul’s speed after being pulled 3.0 m is approximately 1.96 m/s.

Explanation:

Solving this problem involves understanding the work-energy theorem and forces. First, let's calculate the work done. The work done by the force Susan applies (W1) is the product of the tension (T), the distance (d), and the cosine of the angle (θ). W1 = T * d * cos(θ) = 30N * 3.0m * cos(30) = 77.94J.

Next, the work done against friction (W2) is the product of the frictional force and the distance, which is µmgd. Here, µ is the coefficient of friction (0.20), m (10kg) is the mass of the baby, g (9.8m/s2) is the acceleration due to gravity, and d is the distance (3.0 m). W2 = µmgd = 0.20 * 10kg * 9.8m/s2 * 3.0m = 58.8J.

According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. Therefore, the final kinetic energy (and thus the final speed) of Paul will be the initial kinetic energy plus the net work done on him. His initial speed is assumed to be zero, hence the initial kinetic energy is zero. The net work done on him is W = W1 - W2= 77.94J - 58.8J = 19.14J. Setting this equal to the final kinetic energy, (1/2)mv2, allows us to solve for the final speed, v = sqrt((2 * W)/m) = sqrt((2 * 19.14J)/10kg) = 1.96 m/s approximately.

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A screw can be considered a type of

Answers

Fastener because a fastener is something that connects to objects and usually can come apart but can also be permanent

A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.

Answers

Answer:

The minimum transnational speed is 4.10 m/s.

Explanation:

Given that,

Mass of solid ball = 0.6950 kg

Radius = 0.8950 m

Height = 1.377 m

We need to calculate the minimum velocity of the ball at bottom of the loop to complete the track

Using formula velocity at lower point

$v_(min)=√(5gR)$

Put the value into the formula

$v_(min)=√(5*9.8*0.8950)$

$v_(min)=6.62\ m/s$

We need to calculate the velocity

Using conservation of energy

P.E at height +K.E at height = K.E at the bottom

$mgH+(1)/(2)mv^2=(1)/(2)m(√(5gR))^2$

$v^2=(√(5gR))^2-2gH$

$v^2=(6.62)^2-2*9.8*1.377$

$v^2=16.8352$

$v=√(16.8352)$

$v=4.10\ m/s$

Hence, The minimum transnational speed is 4.10 m/s.

Final answer:

The minimum translational speed the solid ball must have when it is at a height H=1.377 m above the bottom of the loop to successfully complete the loop without falling off the track is approximately 7.672 m/s. This was derived using principles of energy conservation.

Explanation:

The minimum translational speed must be sufficient enough to maintain contact with the track even at the highest point of the loop. Using the principle of energy conservation, the total energy at the height H, assuming potential energy to be zero here, should be equal to the total energy at the highest point of the loop. Here, the total energy at height H will consist of both kinetic and potential energy while at the top of the loop it consists of potential energy only. Setting these equations equal to each other: 0.5 * m * v² + m * g * H = m * g * 2R Solving the above equation for v:v = √2g (2R-H). Substituting known values henceforth gives us √2*9.81*(2*0.895-1.377) = 7.672 m/s. Hence, the ball must have a minimum translational speed of approximately 7.672 m/s at height H to complete the loop without falling.

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Help me with my physics, please

Answers

The right answer would be

-20t+ 80

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.

Answers

Answer:

av=0.333m/s, U=3.3466J

$v_(A2)=-1.333m/s,\n v_(B2)=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\n\nP=mv\n\n\therefore m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_(B2)-v_(A2)$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_(B2)=v_(A2)=v_2\n\n2* 2+10* 0=2v_2+10v_2\n\nv_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_(el,1)=K_2+U_(el,2)\n\n#Springs \ in \ equilibrium \ before \ collision\n\nU_(el,2)=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\n\nU_(el,2)=0.5* 2* 2^2-0.5(2+10)(0.333)^2\n\nU_(el,2)=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

$2*2+10*0=2v_A_2+10v_B_2\n\n2=v_A_2+5v_B_2\n\n#Eqtn 2:\nv_A_1-v_B_1=v_(B2)-v_(A2)\n\n2-0=v_(B2)-v_(A2)\n\n2=v_(B2)-v_(A2)\n\n#Solve \ to \ eliminate \ v_(A2)\n\n6v_(B2)=2.0\n\nv_(B2)==0.667m/s\n\n#Substitute \ to \ get \ v_(A2)\n\nv_(A2)=(4)/(6)-2=1.333m/s$

In this example the motion is in a vertical circle. Passengers in a carnival ride travel in a circle with radius 5.0 m. The ride moves at a constant speed and makes one complete circle in a time T=4.0s. What is the acceleration of the passengers? If the ride increases in speed so that T=3.0s, what is arad? (This question can be answered by using proportional reasoning, without much arithmetic.)

Answers

Answer:

a. $12.3m/s^(2)$

b. $21.93m/s^(2)$

Explanation:

From the data given, the radius is 5.0m, and the time taken to complete one circle is 4.0secs

Since the motion is in a circular part, we can conclude that the total distance covered in this time is given as circumference of the circle.

which is expressed as

$Distance=2\pi R$

To determine the speed, we use the equation

$speed=(distnce)/(time)\n Speed=(2\pi R)/(time)\n speed=(2\pi*5 )/(4)\n Speed=7.85m/s$

The acceleration as required is expressed as

$a=(v^(2))/(r)\n a=(7.85^(2))/(5)\n a=12.3m/s^(2)$

if the speed increase and it takes 3secs to complete one circle, the speed is

$speed=(distnce)/(time)\n Speed=(2\pi R)/(time)\n speed=(2\pi*5 )/(3)\n Speed=10.47m/s$

and the acceleration becomes

$a=(v^(2))/(r)\n a=(10.47^(2))/(5)\n a=21.93m/s^(2)$

The acceleration of the passengers in the vertical circle carnival ride is 19.6 m/s^2. When the time taken to complete one circle is 3.0 s, the new acceleration is 26.13 m/s^2.

The acceleration of the passengers can be determined using the centripetal acceleration formula, which is given by a = v^2 / r.

In this case, the velocity v can be found by dividing the circumference of the circle (2πr) by the time taken to complete one circle (T). The radius r is given as 5.0 m. Plugging in the values, we have:

a = (v^2) / r = ((2πr / T)^2) / r = (4π^2r) / T^2 = (4π^2 * 5.0) / 16.0 = 19.6 m/s^2

To find the new acceleration when the time taken to complete one circle is 3.0 s, we can use the proportional reasoning to determine the relationship between the two accelerations. Since the time is inversely proportional to the acceleration, when T is 3.0 s, the new acceleration arad can be found using the equation:

arad / 19.6 = 4.0 / 3.0

Simplifying the equation, arad = (19.6 * 4.0) / 3.0 = 26.13 m/s^2

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