answer

watching

views

27 Nov 2019

You are swinging your younger sister in a horizontal circle ofradius 0.80 m. Her mass is 21 kg and you arrange it so she makesone revolution every 1.5 s.

a) What is the magnitude and direction of the force that must beexerted by you on her? (Model her as a point particle.)
magnitude N?
direction Â°? above/below the horizontal?
(b) What is the magnitude and direction of the force she exerts onyou?
magnitude N?
direction Â°? above/below the horizontal

answer

watching

views

For unlimited access to Homework Help, a Homework+ subscription is required.

You have 0 free answers left.
Get unlimited access to 3.8 million step-by-step answers.

Get unlimited access

Already have an account? Log in

Nelly StrackeLv2

23 Jan 2019

Unlock all answers

Get 1 free homework help answer.

Already have an account? Log in

Related questions

1. A physical therapy patient contracts her biceps muscle, andexerts a horizontal force of 180 N on the spring shown in thefigure at right. Assume the forearm rotates at the elbow. a) Drawan extended force diagram of the forearm (everything from the elbowto the tip of the fingers) showing the pivot point and all of theforces acting on the forearm where they actually are applied. Youwonât know the magnitudes (or even the directions) of these forcesuntil you finish the whole problem but just draw reasonable vectorsfor now.
b) Now draw a force diagram with the forearm as a dot. c) Use thefact that the forearm is perfectly stationary at all times to findthe horizontal force that her biceps exerts on her forearm AND theforce exerted by the bone of the upper arm (shown below the biceps)on the forearm (weight = 30N). Note that muscles can only PULL inthe direction along the muscle, but that ligaments (like the onesin the elbow that connect the upper arm with the forearm) can exertforces in any direction. d) Briefly explain in complete Englishsentences (i.e., not just equations) why the bone of the upper armexerts such a large force on the forearm. e) What is the advantageof having the point where the biceps is attached to the forearm soclose to the elbow? What is the disadvantage of having it attachedso close?

2. You are trying to roll your bicycle wheel (of mass m andradius R) over a curb (of height h). To do this, you apply ahorizontal force F in the direction of desired motion. What is theminimum magnitude of the force that will successfully raise thewheel onto the curb when the force is applied (a) at the center ofthe wheel, and (b) at the top of the wheel? (c) In which case isthe magnitude of the force the least?

3. A hay bale is being dragged along a horizontal surface withconstant speed by a force F. The hay bale is 0.6 m high, 0.3m wide,and 0.9 m long (the side the force is applied to, perpendicular tothe direction of motion). The coefficient of kinetic friction is0.33. (a) Find the magnitude of the force, F. (b) Find the height,h, at which the bale will begin to tip over if the force is appliedat that height.

1.

While helping a friend move, you push a 11.5 kg dolly with a forceof 11.0 N. Assume there are no other forces acting on the dolly. Ifthe dolly starts at rest, how far does it move in 3.50 s?
1 m

2.

At a lumberjack competition an axe with a mass of 1.25 kg is thrownat a target with a speed of 41 m/s. When it hits the target, itpenetrates to a depth of 0.055 m.
(a) What was the average force exerted by the target on theaxe?
1 N

(b) If the mass of the axe is halved, and the force exerted by thetarget on the axe remains the same, by what multiplicative factordoes the penetration depth change?
2
It doubles. It increases by a factor of the square root of two. Itdoes not change. It decreases by a factor of the square root oftwo. It halves.
Your work in question(s) will also be submitted or saved.
3

Your 1500 kg car pulls a 540 kg trailer away from a stop sign withan acceleration of 2.00 m/s2. (Note the answers should be given inkN.)
(a) What is the net force exerted by the car on the trailer?
1 kN

(b) What force does the trailer exert on the car?
2 kN

(c) What is the net force acting on the car?
3 kN

4.

Two boxes sit next to, and touching, each other on a frictionlesshorizontal surface. The lighter box has a mass of 5.31 kg and theheavier box has a mass of 7.29 kg.
(a) Find the contact force between these boxes when a horizontalforce of 6.1 N is applied to the light one.
1 N

(b) If the 6.1 N force is applied to the heavy box instead, is thecontact force between the boxes the same as, greater than, or lessthan the contact force in part (a)?
2

(c) Verify your answer to part (b) by calculating the contact forcein this case.
3 N

5.

A child sits in a wagon and together they have a mass of 25 kg. Thewagon is pulled with a constant speed by a handle that is at anangle of 22Â° above the horizontal. If the normal force exerted onthe wagon is 210 N, what is the force F applied to thehandle?
1 N

6.

An object acted on by three forces moves with constant velocity.One force acting on the object is in the positive x direction andhas a magnitude of 6.7 N; a second force has a magnitude of 4.5 Nand points in the negative y direction. Find the direction andmagnitude of the third force acting on the object.
1 N
2Â° (counterclockwise from the +x axis)

7.

A chain is used to lift a 1200 kg car vertically upward with anacceleration of 0.90 m/s2. What is the tension in the chain?
1 N

8.

A 22.5 kg box (m1) rests on a table.

(a) What is the weight of the box?
1 N
What is the normal force acting on it?
2 N (upward)
(b) A 12.5 kg box (m2) is placed directly on top of the 22.5 kgbox.
Determine the normal force that the table exerts on the 22.5 kgbox.
3 N (upward)
Determine the normal force that the 22.5 kg box exerts on the 12.5kg box.
4 N (upward)

9.

Your friend has a weight of 127 lb when standing on a scale. Youtake the scale to an elevator where you notice your friend's weighton the scale reads 108 lb. What are the direction and magnitude ofthe elevator's acceleration?
Direction 1
Magnitude 2 m/s2

10.

When you lift a package with a force of 82 N, the packageaccelerates upward at a rate of a. If, instead, you lift with aforce of 93 N, the package now has an acceleration of 8 multipliedby a.
(a) Find the weight of the package.
1 N

(b) Find the acceleration a.
2 m/s2

11.

One 3.1 kg paint bucket is hanging by a massless cord from another3.1 kg paint bucket, also hanging by a massless cord, as shownbelow.

(a) If the buckets are at rest, what is the tension in eachcord?
(lower cord) 1 N
(upper cord) 2 N

(b) If the two buckets are pulled upward with an acceleration of1.30 m/s2 by the upper cord, calculate the tension in eachcord.
(lower cord) 3 N
(upper cord)5

12)
Your friend has a pair of fuzzy dice with a mass of 1.32 g hangingfrom his rear-view mirror.
(a) When he accelerates from a stoplight, the dice deflect backwardtoward the rear of the car hanging at an angle of 6.18Â° relative tothe vertical. Find the tension in the string holding thedice.
N

(b) At what angle to the vertical will the tension in the string betwice the weight of the dice?
Â°

A boy in a wheelchair (total mass 50.5 kg) has a speed of 1.40 m/s at the crest of a slope 2.40 m high and 12.4 m long. At the bottom of the slope, his speed is 6.50 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

2 Answers

Relevance

electron1
Lv 7

4 years ago

There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchair’s weight that is parallel to the hill and the force the boy exerts. The net force on the wheelchair is equal to the sum of these two forces minus the friction force. Let’s use the following equation to determine the acceleration.

vf2 = vi^2 + 2 * a * d

6.5^2 = 1.4^2 + 2 * a * 12.4

24.8 * a = 40.29

a = 40.29 ÷ 24.8

This is approximately 1.62 m/s^2.

Net force = 50.5 * 40.29 ÷ 24.8 = 2034.645 ÷ 24.8

This is approximately 82 N.

Force parallel = 50.5 * 9.8 * 2.40 ÷ 12.4 = 1187.76 ÷ 12.4

This is approximately 95.8 N. This force and the force that the boy exerts are the forces that cause the wheelchair to accelerate.

Net force = F + (1187.76 ÷ 12.4) – 41

Set the two net forces equal to each other.

F + (1187.76 ÷ 12.4) – 41 = 2034.645 ÷ 24.8

F = 41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)

F is approximately 27.3 N

Work = [41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)] * 12.4 = 337.9625 J
JupiterSky
Lv 6

4 years ago

His change in kinetic energy is due to gravitational- and his own work,

while the friction does negative work.

∆K = ∆U +∆W + ∆Wf

½m(Vf²−Vi²) = mgh + ∆W + Ff*d

∆W = ½m(Vf²−Vi²) − mgh − Ff*d

∆W = ½*50.5(6.5²−1.4²) − (50.5*9.8*2.4) − (−41*12.4)

∆W = 508 J

Still have questions? Get your answers by asking now.

Next Question

---

asked in Science & MathematicsPhysics · 8 years ago

What is the spring constant? please help me!!!!?

A block of mass 487 g is pushed against the
spring (located on the left-hand side of the
track) and compresses the spring a distance
4.7 cm from its equilibrium position (as shown
in the ﬁgure below). The block starts from
rest, is accelerated by the compressed spring,
and slides across a frictionless horizontal track
(as shown in the ﬁgure below). It leaves the
track horizontally, ﬂies through the air, and
subsequently strikes the ground.
acceleration = 9.81
 
A) What is the spring constant?
B)What is the speed v of the block when it leaves
the track?
C) What is the total speed of the block when it
hits the ground?
...Show more

Update:

Height Above ground = 1.9 m
Lands = 4.59m away

What is the spring constant? please help me!!!!?

3 Answers

Relevance

Jánošík
Lv 7

8 years ago

Favorite Answer

I'm sure there is some crucial information in "the figure below" !!!

Edit:

OK... now we got something to work with ... ;-)

Time needed to fall a vertical distance of 1.9 m is:

t = √( 2h / g )

t = √[ 2×(1.9 m) / (9.81 m/s²) ]

t = 0.6224 s

In that time, the block must travel a horizontal distance of 4.59 m, so the horizontal velocity is:

r = v×t

(4.59 m) = v×(0.6224 s)

v = 7.37 m/s < - - - - - - - - - - - - - - - - - answer B

That is also the speed of the block as it leaves the spring.

Its kinetic energy at that time is:

KE = m×v²/2

KE = (0.487 kg)×(7.3749 m/s)²/2

KE = 13.244 J

That is also the elastic potential energy stored in the spring at maximum compression. So the spring constant is:

KE = PEe = k×d²/2

(13.244 J) = k×(0.047 m)²/2

k = 12 N/m < - - - - - - - - - - - - - - - - - - answer A

The gravitational potential energy of the block on the table is:

PEg = m×g×h

PEg = (0.487 kg)×(9.81 m/s²)×(1.9)

PEg = 9.077 J

The total energy of the block on the table is:

TE = PEg + PEe

TE = (9.077 J) + (13.244 J)

TE = 22.32 J

When hitting the ground, all that energy is converted to kinetic energy, so

TE = KE = m×v²/2

(22.32 J) = (0.487 kg)×v²/2

v = 9.57 m/s < - - - - - - - - - - - - - - - - - - answer C

...Show more
Technobuff
Lv 7

8 years ago

There must be a height above ground and a distance from the end of the track on the "as shown in the figure below".??
STALKER
Lv 4

8 years ago

use f=kl

where k is the spring constant and l is extension from the natural length.

k=f/l

since you have mass in grams. use w=mg

w=0.487 x 9.8 = 4.77N

Now use the equation,

k=4.77/4.7

k=1.012

Note: my answer could be wrong since I do not know the springs original length from the diagram shown. But, the method of working out spring constant is same.

...Show more

Still have questions? Get your answers by asking now.

Next Question

ひいらぎ

Lv 5

asked in Science & MathematicsPhysics · 1 decade ago

Physics Momentum, Rolling?

Pat builds a track for his model car out of wood. The track is 10.00 cm wide and 3.00 m long, and is solid. The runway is cut such that it forms a parabola by the equation y=(x - 3)^2/9. Locate the horizontal position of the center of gravity of this track.
 
A washing machine load in the "spin" cycle, (much like a centrifuge used to separate fluids), can be modeled as a cylinder which has zero density toward the center, and starting from some inner radius where the clothes (or fluids) begin, increases in density as you move radially outward. The washing machine has density of clothes (in SI units) of p=14r+5, a height of 0.51 m, and only has clothes present from inner radius 0.2m out to 0.7m from the center.
-What is the moment of Inertia of the wash load?
-What is the power of motor that drives the clothes from rest up to 4 rev/sec, in 14 seconds if the basin has moment of Inertia of 4 kg*.m2
...Show more

Physics Momentum, Rolling?

1 Answer

Relevance

Anonymous

1 decade ago

Favorite Answer

I can't visualize the first problem.

The second: Moment of inertia is I = m r^2. In a shell of thickness dr there is a mass of p * 2 * pi * r * h * dr. The moment of inertia of the shell is 2 pi h p r^3 dr. Now integrate this from r = 0.2m to r = 0.7m.

Kinetic energy is 1/2 m v^2. In rotations the moment of inertia is like the mass and the angular velocity is like the velocity, so rotational energy is 1/2 I omega^2. Omega is in rad/s. The total inertia of the system is the sum of the inertia of the clothes and the basin. Divide the energy by the time needed to get the power.

...Show more

Still have questions? Get your answers by asking now.

Next Question

asked in Science & MathematicsPhysics · 1 decade ago

Physics:You and your friends push a 75-kg...Help!?

You and your friends push a 75-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070.
 
(a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 4.9 m/s2 over a distance of 1.8 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 50 cm. What is the angle of inclination of the slide? __degree
 
(b) At the maximum height the pig turns around and begins to slip down the slide, how fast is it moving when it arrives at the low end of the slide? __m/s
...Show more

Physics:You and your friends push a 75-kg...Help!?

1 Answer

Relevance

Anonymous

1 decade ago

Favorite Answer

♣the distance the pig was pushed up is

s=0.5*a*t^2, where s=1.8m, a=4.9m/s^2, t is time of push-up;

speed of the pig gained after s=1.8m is v=a*t; thus v^2=2as;

♠ kinetic energy of the pig at the moment it’s released after push-up is

E=0.5*m*v^2, which is enough for the pig to move by itself L distance more along the ramp, where L=h/sin(b), h=50cm=0.5m, b is angle in question;

♦pig’s potential energy at vertex (when stopped) is E1=mgh;

the energy E2=E-E1 is gone on friction, that is E2=f*L is work of friction, where

f=μ*w1, w1=mg*cos(b) is component of pig’s weight normal to the aluminum ramp; or;

f= μ*mg*cos(b);

♦Thus E2=E-E1; or; f*L = 0.5*m*v^2 –mgh; or;

(μ*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as –mgh; or;

μ*gh*cos(b)/sin(b) =as –gh, hence cot(b) =(as –gh)/(μ*gh), hence

(a); b=atan(μ*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5°;

♥ pig’s potential energy above the low end of the ramp is

W= mg*(s*sin(b) +h); pig’s ramp position is p=s +L = s + h/sin(b);

Pig’s final kinetic energy is W1=W-W2, where W2=f*p is work of friction;

and W1=0.5*m*v^2; thus

0.5*m*v^2 = mg*(s*sin(b) +h) – μ*mg*cos(b)*(s + h/sin(b));

v^2 =2g*(s*sin(b) +h)*(1 –μ*cot(b)) =2.575;

(b); v =1.605 m/s;

...Show more

Still have questions? Get your answers by asking now.

+100

Join Yahoo Answers and get 100 points today.

Weekly leaderboard

Home

Homework Help 3,900,000

Physics 110,000

Unlock all answers

Related textbook solutions

Physics: Principles with Applications

Related questions

2 Answers

What is the spring constant? please help me!!!!?

3 Answers

Physics Momentum, Rolling?

1 Answer

Physics:You and your friends push a 75-kg...Help!?

1 Answer

Trending Questions

Weekly leaderboard