You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height h from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 2.0 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 30.0 m/s when they reach the bottom of the ramp. You determine that for a 77.0 kg skier with good form, friction and air resistance will do total work of magnitude 4000 J on him during his run down the slope.

Required:
What is the maximum height h for which the maximum safe speed will not be exceeded?

Answers

Answer 1

Answer:

h = 40.37 m

Explanation:

We will apply the law of conservation of energy to the skier in this case, as follows:

\(Energy\ of\ skier\ at\ the\ gate = Energy\ of\ Skier\ at\ the\ end\\P.E + K.E_{i} = K.E_{f} - W_{friction}\\mgh + \frac{1}{2}mv_{i}^2 = \frac{1}{2}mv_{f}^2 - W_{friction}\\\\mgh = \frac{1}{2}m(v_{f}^2-v_{i}^2) - W_{friction}\)

where,

m = mass of skier = 77 kg

g = acceleration due to gravity = 9.81 m/s²

vf = final speed = 30 m/s

vi = initial speed = 2 m/s

W_friction = Work done by friction and air resistance = 4000 J

Therefore,

\((77\ kg)(9.81\ m/s^2)h = \frac{1}{2}(77\ kg)[(30\ m/s)^2-(2\ m/s)^2] - 4000\ J\\\\h = \frac{34496\ J - 4000\ J}{755.37\ N}\\\\\)

h = 40.37 m


Related Questions

Which of the following is a problem with the estimated age of the universe?

There are stars in our galaxy that appear to be over 15 billion years old.

The universe is shrinking, which makes it is impossible to tell how old it is.

There should be more supergiants if the universe is really about ten billion years old.

There are planets in other galaxies that are less than five billion years old.

Answers

Answer:

There are stars in our galaxy that appear to be over 15 billion years old.

Explanation:

The age of the universe is the time past since the big ban event as based on observations of the early state of the universe. The comic problem with the universe is that it's estimated to be older since the time of the big bang. It also represents a contrast with objects in the galaxies such as stars, planets that could not have been present.

The diagram below shows the velocity vectors for two cars that are moving
relative to each other.
45 m/s west
25 m/s east
Car 1.
Car 2
From the frame of reference of car 2, what is the velocity of car 1?
OA. 70 m/s east
B. 20 m/s west
OC. 70 m/s west
OD. 20 m/s east
SUBMIT

Answers

The velocity of the car 1 can be seen from the calculation as 20 m/s West

What is relative motion?

A coordinate system or point of view used to observe motion is known as a frame of reference. It can be used as a guide when describing an object's position, speed, and acceleration. Different frames of reference may result in various motion observations.

The relative velocity is the velocity of an object or observer as observed from a particular frame of reference.

We can see that the velocity of the car 1 is;

45 m/s - 25 m/s

= 20 m/s West

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If
\( \binom{13}{7} \)
N has a half-life of about 10.0 min, how long will it take for 20 g of the isotope to decay to 1.9 g? [2 marks] 2. Radon-22​

Answers

Answer:

the time it will take the element to decay to 1.9 g is 34.8 mins.

Explanation:

Given;

half life of Nitrogen, t = 10 min

initial mass of the element, N₀ = 20 g

final mass of the element, N = 1.9 g

The time taken for the element to decay to final mass is calculated as follows;

time (min)                        mass remaining

0 ----------------------------------20 g

10 mins ------------------------- 10 g

20 mins ------------------------- 5 g

30 mins -------------------------- 2.5 g

40 mins --------------------------- 1.25 g

Interpolate between 2.5 g and 1.25 to obtain the time for 1.9 g

30 min ------------------------- 2.5 g

x ----------------------------------- 1.9 g

40 min -------------------------- 1.25 g

\(\frac{30 - x}{40- 30} = \frac{2.5 - 1.9}{1.25 - 2.5} \\\\-1.25(30-x) = 6\\\\-37.5 + 1.25x = 6\\\\1.25x = 6+37.5\\\\1.25x = 43.5\\\\x = \frac{43.5}{1.25} \\\\x = 34.8 \ mins\)

Therefore, the time it will take the element to decay to 1.9 g is 34.8 mins.

Simple machines make work easier. You must left a move this load of Saul for your parents. It is heavy you’re using to simple machines when you use this wheel barrow wight simple machine included in the wheelbarrow makes it easier to move the song from one place to another.

Answers

The use of the wheel barrow in this case makes it a simple machine

What is the simple machine?

A wheelbarrow is an illustration of a straightforward device that can simplify labour by lowering the force needed to move a large load. The wheel and axle and the lever are two basic machines that are used.

The wheelbarrow's wheel is made of an axle and wheel. The friction between the wheel and the ground is decreased by the wheel's rotation around the axle. It can also be seen as one of the classes of the lever.

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a tennis ball is swung in a horizontal circle by a string. if the string is broken, which picture correctly shows the path the ball will take?

Answers

Initially rolling inside a horizontal circle five meters above the ground at a speed of two meters per second, the ball breaks the string as it approaches true North. Though it is falling, it is still heading in the direction

Why does string theory defy physics?

The lower string, however, is the one that snaps if it is forcefully yanked. The prevailing theory holds that the mass's inertia prevents it from moving and, as a result, from pulling on the top string.

Which string will finally snap?

If both strings are equally strong, the top string will break sooner. However, a change in tension from the lower string takes a moment to transfer because of the inertia of a hanging object.

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The reliability or reproducibility of a measurement is its _____

Answers

The degree of data stability when the measurement is replicated under identical circumstances is known as reproducibility or reliability.

What exactly are repeatability and reproducibility?

Reproducibility determines how an entire study an experiment can be replicated, whereas repeatability assesses the variation in data made by a single equipment or human under similar circumstances.

What makes repeatability crucial?

Science needs reproducibility because it enables more in-depth investigation, and replication validates our findings. There are several investigations and experiments, which result in a wide range of variables, unforeseen, and things that are either outside your influence or you cannot guarantee.

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A force of 25 newtons moves a box a distance of 4 meters in 5 seconds. The work done on the box is_ No, and the power is_Nm/s

Answers

Explanation:

workdone = force x distance

workdone= 25 x 4

workdone = 100 joules ( the unit for workdone is joules or Nm )

POWER = workdone/ time

power = 100/5

power = 20 watts ( the unit for power is watts or Nm/s )

Two horizontal forces act on an object. One force has a magnitude of 75.0 N and is directed due 30.0° south of east. The other force has a magnitude of 55.0 N and is directed due 70.0° north of west. What is the magnitude and direction of the sum of the two force vectors. Select one: 122.4 N, 43.2° north of east 130.0 N, 40.0° south of east 46.6 N, 17.1° south of east 48.3 N, 17.1° north of east 20.0 N, 40.0° north of east

Answers

Explanation:

70 degrees North of west = 110 degrees

30 degrees southof east = - 30 degrees:

Vertical components added together =

  75 (sin -30)   +    55 sin 110  =  14.18  N

Horizontal components added together =

 75 cos (-30)   + 55  cos 110    = 46.14  N

Magnitude =  sqrt ( 14.18^2 + 46.14^2 ) = 48.3   N

   direction = arctan ( 14.18 / 46.14) =  ~ 17.1 North of East

s the gravitational force greater between the objects in Pair 1 or Pair 2? Explain why. (Picture shown below)

s the gravitational force greater between the objects in Pair 1 or Pair 2? Explain why. (Picture shown

Answers

Answer:

Pair 1

Explanation:

The gravitational force greater between the objects in Pair 1 because in Pair 2 the objects are far apart and in pair 1 the object are more closer to each other.

So, Gravitational force directly proportional to distance

Thus, Increase in distance = Increased in gravitational force

-TheUnknownScientist

An aeroplane is circling above an airport in a horizontal circle at a speed of 400 kmh-1.The banking angle of the wings is 20.What is the radius of the circular path?

Answers

Answer: the radius of the circular path is approximately 1637.58 meters.

Explanation:

The centripetal force acting on the airplane is provided by the component of the gravitational force that acts towards the center of the circular path. This component is given by:

F_c = m * g * tan(banking angle)

Where:

F_c is the centripetal force

m is the mass of the airplane

g is the acceleration due to gravity

tan(banking angle) is the tangent of the banking angle

Now, the centripetal force is also given by the formula:

F_c = (m * v^2) / r

Where:

v is the speed of the airplane

r is the radius of the circular path

Equating the two expressions for F_c, we get:

(m * g * tan(banking angle)) = (m * v^2) / r

Canceling out the mass (m) on both sides of the equation, we have:

g * tan(banking angle) = v^2 / r

Solving for r, we get:

r = (v^2) / (g * tan(banking angle))

Substituting the given values:

v = 400 km/h = 400,000 m/h

g = 9.8 m/s^2

banking angle = 20°

Converting the speed to m/s:

v = 400,000 m/h * (1/3600) h/s = 111.11 m/s

Converting the banking angle to radians:

banking angle = 20° * (π/180) rad/° = 0.3491 rad

Now, substituting the values into the formula:

r = (111.11^2) / (9.8 * tan(0.3491))

r ≈ 1637.58 meters

Therefore, the radius of the circular path is approximately 1637.58 meters.

Calculate Time
d
12. A vehicle drives a distance of 26000 m at a speed of 65m/s, calculate the time taken for
this journey.
13. A train travels at a speed of 16 m/s and travel a distance of 3200 m, calculate the time it
takes the train to complete this journey.
urs 14. Calculate the time it takes to travel a distance of 672 km at a speed of 96 km/h.
15. A beetle travels at a speed of 0.09 m/s, it travels a distance of 1.08 m before it is caught
in a jar. Calculate the time taken for the beetle to run.
16. Carlisle is a distance of 35 miles away from Lockerbie. If I travelled at a constant speed
5147
deudate the time takon for this journey

Answers

12. The time taken for the journey is 400 s

13. The time taken for the train is 200 s

14. The time taken is 7 h

15. The time taken for the beetle is 12 s

16. The time taken for the journey is 0.0068 h

How do i determine the time taken?

The time taken in each case as given by the question can be obtain as follow:

12. The time taken for the journey

Distance traveled = 26000 mSpeed = 65 m/s Time taken =?

Time taken = Distance / Speed

Time taken = 26000 / 65

Time taken = 400 s

13. The time taken for the train

Distance traveled = 3200 mSpeed = 16 m/s Time taken =?

Time taken = Distance / Speed

Time taken = 3200 / 16

Time taken = 200 s

14. The time taken to travel

Distance traveled = 672 kmSpeed = 96 Km/h Time taken =?

Time taken = Distance / Speed

Time taken = 672 / 96

Time taken = 7 h

15. The time taken for the beetle

Distance traveled = 1.08 mSpeed = 0.09 m/s Time taken =?

Time taken = Distance / Speed

Time taken = 1.08 / 0.09

Time taken = 12 s

16. The time taken for the journey

Distance traveled = 35 milesSpeed = 5147 mile per hourTime taken =?

Time taken = Distance / Speed

Time taken = 35 / 5147

Time taken = 0.0068 h

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You use the equation for finding potential energy (PE=mgh) to find that a marble with a mass of20g held in the air has 0.2 J of potential energy. You then drop the marble. Work the follow outon a piece of paper carefully showing all work:(a) Since you know that the Conservation of Mechanical Energy applies(TE=KE+PE), use the fact there is no kinetic energy when the marble is heldstill in the air to find the total energy that the marble has when it's dropped.(b) Since you know that the total energy is conserved so that the total energy ofthe marble that you found in (a) equals the total energy of the marble rightbefore it hits the ground, write the equation for TE of the marble right beforeit hits the ground subbing in the value for the TE of the marble found in (a).

Answers

Given:

The potential energy of the marble when it is held in the air, PE=0.2 J

The mass of the marble, m=20 g

The initial kinetic energy of the marble, KE=0 J

To find:

(a) The total energy of the marble.

(b) The total energy of the marble before it hits the ground.

Explanation:

(a)

The total mechanical energy of an object is the sum of its potential and kinetic energy.

Thus the total energy of the marble when it is dropped is

\(\begin{gathered} TE=PE+KE \\ =0.2+0 \\ =0.2\text{ J} \end{gathered}\)

(b)

The total energy of an object will always remains the same as stated by the law of conservation of energy.

Thus the total energy of the marble will be the same as it energy when it was dropped.

Therefore, the total energy of the marble right before it hits the ground is \

\(TE=0.2\text{ J}\)

Final answer:

a) The total energy of the marble when it is dropped is 0.2 J

b) The total energy of the marble right before it hits the ground is 0.2 J

A car travels from point A to B in 3 hours and returns back to point A in 5 hours. Points A and B are 150 miles apart along a straight highway. Calculate: a) Total distance and total displacement (in mile and meter) b) Average speed and Average velocity (in mile/hr and m/s​

Answers

The total distance covered by the car is 300 miles.

The total displacement covered by the car is zero.

The average speed of the car is 17.88 m/s.

The average velocity of the car is also zero.

Distance between the points A and B, d = 150 miles

Time taken by the car to travel from A to B, t₁ = 3 hours

Time taken by the car to travel from B to A, t₂ = 5 hours

a) Given that the car travelled from A to B and then back to A.

Therefore, the total distance covered by the car is,

Distance = 2 x d

Distance = 2 x 150

Distance = 300 miles

Since the car is travelling from A to B and then returning back to the initial point A, the total displacement covered by the car is zero.

b) The speed with which the car travelled from A to B is,

v₁ = d/t₁

v₁ = 150/3

v₁ = 50 miles/hr

v₁ = 22.35 m/s

The speed with which the car travelled from B to A is,

v₂ = d/t₂

v₂ = 150/5

v₂ = 30 miles/hr

v₂ = 13.41 m/s

Therefore, the average speed of the car is,

v = (v₁ + v₂)/2

v = (22.35 + 13.41)/2

v = 17.88 m/s

As, the total displacement of the car is zero, the average velocity of the car is also zero.

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Can a machine multiply both force and speed at same time??​

Answers

Answer:

No

Explanation:

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

An intrepid hiker reaches a large crevasse in his hiking route. He sees a nice landing ledge 60.0 cm below his position but it is across a 2.3 m gap. He spends 1.2 s accelerating horizontally at 5.92 m/s2 [right] in an attempt to launch himself to the safe landing on the far side of the gap. Does he make it?

Answers

The hiker made it to a safe landing on the other side of the gap after travelling horizontally at 2.49 m.

What is the time motion from the vertical height?

The time taken for the hiker to fall from the given height is calculated as follows;

h = vt + ¹/₂gt²

where;

v is the vertical velocity = 0t is the time of motiong is acceleration due to gravityh is the height of fall

h = ¹/₂gt²

t = √(2h/g)

t = √[(2 x 0.6) / (9.8)]

t = 0.35 seconds

The horizontal velocity of the hiker during the period of acceleration is calculated as follows;

Vₓ = at

Vₓ = (5.92 m/s²) x (1.2 s)

Vₓ = 7.104 m/s

The horizontal distance travelled during the time period of 0.35 seconds;

X = Vₓt

X = 7.104 x 0.35

X = 2.49 m

Thus, the hiker made it to a safe landing on the other side of the gap which is 2.3 m wide and smaller to his horizontal displacement of 2.49 m.

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An 4.0-kg fish pulled upward by a fisherman rises 1.9 m in 2.4 s, starting
from rest. Assuming the acceleration is constant, find the magnitude
and direction of the net force acting on the fish during this interval.

Answers

Answer:

2.64N  

Explanation:

Force = mass * acceleration

Given

mass = 4kg

distance = 1.9m

Time t = 2.4s

Get the acceleration using the equation of motion

S = ut + 1/2at²

1.9 = 0 + 1/2a(2.4)²

1.9 = 5.76a/2

1.9 = 2.88a

a = 1.9/2.88

a = 0.66m/s²

Get the magnitude of the force

Force = 4 * 0.66

Force = 2.64N

Hence the net force acting on the fish is 2.64N  

Which of the following relationships is correct?
2 points
1 N = 1 kg
1 N = 1 kg·m
1 N = 1 kg·m/s
1 N = 1 kg·m/s2

Answers

1 N = 1 kg•m is the answer

PLEASE Which list places the three types of radiation in order from the one that is
most difficult to block to the one that is easiest to block?
A. Gamma, alpha, beta
B. Alpha, beta, gamma
C. Alpha, gamma, beta
D. Gamma, beta, alpha

Answers

Answer:

D

Explanation:

Gamma, beta, alpha

If right, please mark me brainliest

The types of radiation from the most difficult to easiest to block is Gamma, beta, alpha.

What is radiation and explain its types?

         The energy from the source that travel with the speed of light to the space is known radiation. There are three types of radiation:

they are:

Alpha particlesBeta particlesGamma rays

   Alpha particles is a particle which consists of 2 proton and 2 neutron.

The alpha particles can  easy to block.

eg: Helium nucleus

   Beta particles is a ray or radiation that emits electron and positron during the process of beta decay.

   Gamma rays is a radiation which is the penetration of electromagnetic radiations during radioactive decay. These radiations were very difficult to block.

So, Option D is the correct answer.

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Sound wave A is moving through a medium that is decreasing temperature.
Sound wave B is moving from a liquid to a gas medium
Which statement can be made about both waves?

1. They will speed up

2. They will increase the
density of their medium

3. They will slow down

4. They will decrease the density of their medium

Answers

The satement that can be made about sound wave A and sound wave B is, they will slow down.

Relationship between sound wave and temperature

The relationship between sound waves and temperature is given by the following formula;

\(v = \sqrt{\frac{\gamma RT}{M} }\)

The speed of sound wave increases with increase in temperature, and vice versa.

Speed of sound wave in liquid and gaseous medium

Sound wave is mechanical wave, because it requires material medium for its propagation. Sound will travel faster in liquid medium than gaseous medium because of number of molecules per unit volume.

Thus, the satement that can be made about sound wave A and sound wave B is, they will slow down.

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A 30-microfarad capacitor is charged to 90 v and then connected across an initially un charged capacitor of unknown capacitanc c .if the final potintial difference across the 30- microfarad capacitor is 20 v sitemin c

Answers

Answer:

The value  is   \(  C_u   =   5 *10^{-6} F\)

Explanation:

From the question we are told that

   The voltage of the capacitor is  \(V  =  90 \  V\)

   The  capacitance of the capacitor is  \(C  =  30 \ \mu F =  30 *10^{-6} \  F\)

   The  final voltage is  \(V_a =  20 \  V\)

Since the capacitors are connected in parallel we have that

      \(Q_u  =  Q'_u  +  Q\)

Where  \(Q_u\) is the charge of the known capacitor before it is connected to the known capacitor

     \(Q_u' \) is the charge of the known capacitor after it is connected

       \(Q  \) is the charge of the unknown capacitor

Also the potential across the two capacitors will be the same (except for a loss due heat (it been converted to heat ))

So

            \(CV  =  CV_a +  C_u *  V_a\)

=>          \(  C_u *  V_a   =  CV - CV_a \)

=>          \(  C_u   =   \frac{CV - CV_a }{ V_a } \)

=>          \(  C_u   =   \frac{ [30*10^{-6} *90] - [30*10^{-6} * 20] }{ 20 } \)

=>        \(  C_u   =   5 *10^{-6} F\)

Rhea is driving north in a straight line. After driving for 2.4 kilometers, she turns west, and drives for 3.1 km. At the end of her drive, what is the magnitude of her displacement vector? To find the magnitude of a resultant vector, use the Pythagorean theorem: a² + b² = c² A) 3.9 km B) 5.5 km (this one is incorrect) C) 2.8 km D) 2.3 km

Answers

Option A) 3.9 km is the correct answer. the magnitude of Rhea's displacement vector is approximately 3.92 km.

In order to find out the magnitude of Rhea's displacement vector, we have to add up all of the displacement vectors.

Then we can use the Pythagorean theorem to calculate the magnitude of the resultant vector.

Since Rhea is first driving north for 2.4 km and then west for 3.1 km, we can represent her displacement vectors as follows: Δx = 0 km and Δy = 2.4 km for the first vector, and Δx = -3.1 km and Δy = 0 km for the second vector.

We can then add these vectors together by adding their components: Δx = 0 km + (-3.1 km) = -3.1 km and Δy = 2.4 km + 0 km = 2.4 km.

This gives us a resultant vector of -3.1 km east and 2.4 km north.

Using the Pythagorean theorem, we can find the magnitude of this vector: \(\sqrt{(\(-3.1 km)^{2} + (2.4 km)^{2} ) } = \sqrt{(9.61 + 5.76) km^{2} } = \sqrt{15.37 km^{2} } \approx 3.92 km.\)

Therefore, the magnitude of Rhea's displacement vector is approximately 3.92 km.

Therefore, option A) 3.9 km is the correct answer.

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solve this with figure.help me ......​

solve this with figure.help me ......

Answers

Answer:

\( \huge\mathfrak\pink{Hope \: it \: helps}\ \)

solve this with figure.help me ......
solve this with figure.help me ......

Distinguish between the two main types of mechanical waves

Answers

Explanation:

Mechanical waves are waves that requires a medium for their propagation. There are two types of mechanical waves:

Transverse waves Longitudinal waves

Transverse waves are waves that propagates perpendicularly to their source. Examples are electromagnetic waves, ripple on water etc.

Longitudinal waves are waves that propagates parallel to their source. They move in a series of rarefactions and compressions along their path. An example is seismic p-waves

An airplane accelerates at 16 m/s² from rest. If it lifts off from an aircraft carrier after 6.0 s, how far did it go down the runway before lift off?

A. 48m
B. 96m
C. 290m
D. 580m​

Answers

B is the answer
16 x 6.0 = 96
Helping is my pleasure

Given the list of items, select all that are considered to be matter:
music
heat
air
dreams
water
gasoline
love
bacteria
thoughts

Answers

From the given list of items, examples of matter include, heat, air, water, gasoline, and bacteria.

What is matter?

Matter is a substance made up of various types of particles that occupies physical space and has inertia.

A matter must have mass and occupy space.

Examples of matter include the following;

heatairwatergasolinebacteria

Thus, from the given list of items, examples of matter include, heat, air, water, gasoline, and bacteria.

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At the end of the lab all the matter (such as the string, the ball and the books) stayed in the area. The potential energy, that was stored in the ball on the high shelf and suspended weights, as well as the stored mechanical energy of the sling shot, had left the system in the form of sound, energy of motion (kinetic energy) and heat. What type of energy system (open, closed or isolated) is represented by the lab device and why?

Answers

The type of energy system is closed or isolated because the total energy of the system is conserved.

What is principle of conservation of energy?

The law of conservation of energy states that the total energy of an isolated system remains constant; the energy is conserved over time.

When the potential energy, that was stored in the ball on the high shelf and suspended weights, as well as the stored mechanical energy of the sling shot, had left the system in the form of sound, energy of motion (kinetic energy) and heat, the energy of the system has be conserved.

Thus, the type of energy system is closed or isolated because the total energy of the system is conserved.

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Small frogs that are good jumpers are capable of remarkable acceleration. One species of small frog reaches a takeoff speed of 3.7 m/s in 60 ms.
How high can the frog jump given it's acceleration is 62 m/s^2?

Answers

Hi there!

Begin by solving for the time it takes the frog to reach the top of its trajectory. (Vertical velocity = 0 m/s)

\(\large\boxed{v_f = v_i + at}\)

vf = final velocity (0 m/s)

vi = initial velocity (3.7 m/s)

a = acceleration due to gravity (9.8 m/s²)

t = time (s)

Plug in the known values:

\(0 = 3.7 - 9.8t\\\\-3.7/(-9.8) = t = 0.378 s\)

Now, we can use the following kinematic equation:

\(\large\boxed{d = v_it + \frac{1}{2}at^2}\)

d = displacement (m)

vi = initial velocity (3.7 m/s)

a = acceleration (9.8 m/s²)

t = time (0.378 sec)

Plug in the given values:

\(d = 3.7(0.378) - \frac{1}{2}(9.8)(0.378)^2 = \boxed{0.698 \approx 0.7m}\)

how long would it take for a resultant upward force of 100 N to increase the speed of 50 Kg object from 100 m/s to 150 m/s ?

With explanations please....and the very ways..

Answers

Answer:

\(force = mass \times acceleration \\ 100 = 50 \times a \\ a = 2 \: {ms}^{ - 2} \\ from : \: \: v = u + at \\ 150 = 100 + (2 \times t) \\ 50 = 2t \\ time = 25 \: seconds\)

A string is wound around a uniform disc of radius 0.68 m and mass 3.7 kg. The disc is released from rest with the string vertical and its top end tied to a fixed support. calculate the speed of the center of mass when, after starting from rest, the center of mass has fallen 1.2 m. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

Answers

Answer:

3.962

Explanation:

mass = 1.2

gravity g = 9.81

velocity = ?

mgh = 1/2mv² + 1/2(1/2)mv²

= 9.81 x 1.2  = 3/4v²

11.772 x 4 = 3/4v²

47.088 = 3v²

divide through both sides by 3

15.696 = v²

take the square root of both sides

√15.696 = √v²

= 3.6962 = v

the speed is therefore 3.692

A 24.4kg dog is running northward at 2.14m/s, while a 5.53kg cat is running eastward at 3.56m/s. Their 78.5kg owner has the same momentum as the two pets taken together. Find the direction of the owner's velocity. Find the magnitude of the owner's velocity.

Answers

The owner's velocity is in the opposite direction of the combined velocity of the dog and the cat, and its magnitude is approximately 0.916 m/s.

To solve the given problem, we can use the principle of conservation of momentum to find the direction and magnitude of the owner's velocity.

Let's denote the velocity of the dog as v1 (northward), the velocity of the cat as v2 (eastward), and the velocity of the owner as v (unknown).

According to the conservation of momentum, the total momentum before the interaction is equal to the total momentum after the interaction.

The total momentum before the interaction is given by:

Total momentum before = (mass of the dog * velocity of the dog) + (mass of the cat * velocity of the cat) + (mass of the owner * velocity of the owner)

Mass of the dog (m1) = 24.4 kg

Velocity of the dog (v1) = 2.14 m/s

Mass of the cat (m2) = 5.53 kg

Velocity of the cat (v2) = 3.56 m/s

Mass of the owner (m3) = 78.5 kg

Velocity of the owner (v) = unknown

Total momentum before = (24.4 kg * 2.14 m/s) + (5.53 kg * 3.56 m/s) + (78.5 kg * v)

The total momentum after the interaction is zero since the owner has the same momentum as the pets taken together.

Total momentum after = 0

Equating the two expressions:

(24.4 kg * 2.14 m/s) + (5.53 kg * 3.56 m/s) + (78.5 kg * v) = 0

Simplifying the equation:

(52.216 kg·m/s) + (19.6488 kg·m/s) + (78.5 kg * v) = 0

71.8648 kg·m/s + (78.5 kg * v) = 0

Solving for v:

78.5 kg * v = -71.8648 kg·m/s

v = -71.8648 kg·m/s / 78.5 kg

v ≈ -0.916 m/s

Therefore, the direction of the owner's velocity is opposite to the combined velocity of the dog and the cat, and the magnitude of the owner's velocity is approximately 0.916 m/s.

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