Question

Help please

Answer 1

a. Convert 55 inches to centimeters.

55*2.54=137.9 centimeters

b. Convert 5000 pounds to kilograms.

5000*0.454=2270 kilograms

c. Convert 100 miles to kilometers.

100*1.61=161 kilometers

d. Convert 75 liters to quarts.

Answer 2

@Vocaloid

Answer 3

good on a-c

for d) since you are given 1 quart = 0.946 L you need to divide 75L by 0.946 L/q to cancel out units and get q

Answer 4

a. Convert 55 inches to centimeters.

55*2.54=137.9 centimeters

b. Convert 5000 pounds to kilograms.

5000*0.454=2270 kilograms

c. Convert 100 miles to kilometers.

100*1.61=161 kilometers

d. Convert 75 liters to quarts.

75*0.946=70.95 quarts
e. Convert 800 cm to meters.

800*10=8000 meters

Answer 5

for d you need to divide by the conversion factor since you are given the factor for 1 quart not 1 L

Answer 6

for e - if 100cm = 1 m then you need to divide cm by 100 to get m

Answer 7

75/0.946=79.3

Answer 8

good, 79.3 quarts

Answer 9

so 800/10?

Answer 10

yes

Answer 11

800/10=80 meters

Answer 12

oh no, 800/100 sorry

Answer 13

since 1 m = 100 cm

Answer 14

800/100=8 meters

Answer 15

yes

Answer 16

No need for the apology

Answer 17

A diver of mass 75 kg jumps upward off a diving board and into the water. The diving board is 2 m above the water.

a. When the diver jumps, she has a speed of 2.5 m/s. How much kinetic energy does she have?


b. How much gravitational potential energy will she have gained at her highest point? Explain how you found the answer.


c. Based on the diver's speed from part (a), how high above the board will she rise?


d. How fast will the diver be moving when she hits the water?

Answer 18

a) KE = (1/2)mv^2

Answer 19

KE = (1/2)mv^2
KE=(1/2)(75)(2.5)^2
KE=234.38J

Answer 20

1470 m/s

Answer 21

units of potential energy are joules not m/s

Answer 22

GPE = mgh
GPE=(75)(2)(9.8)
GPE=1470J

Answer 23

Sure

Answer 24

I think it's asking for the greatest potential energy after she jumps (largest height) so b) should be equal to the energy from a) since all the energy that goes into jumping is converted into the potential energy increase, so it's also 234.38J

Answer 25

and c is 2 right?

Answer 26

c) is asking for the height with respect to the diving board, so you need to convert the KE from jumping into potential energy according to the equation

KE = mgh

so

234.38J = mgh solve for h

Answer 27

234.38J= mgh
234.38=(75)(9.8)h
h=0.32

Answer 28

good, and since it's a distance units are m

Answer 29

for d) we need to set the total potential energy (including the PE from the climb) into KE

so

mgh = 1/2mv^2, where h is the total height (2 + 0.32)

Answer 30

mgh = 1/2mv^2, where h is the total height (2 + 0.32)
(75)(9.8)(2.32)= 1/2(75)v^2,

Answer 31

good, then solve for v

Answer 32

6.74

Answer 33

6.74

Answer 34

good, since it's velocity units are m/s

Answer 35

At the beginning of a bobsled race, teammates push their bobsled along the track before jumping into it.

a. Before the teammates jump in, they push the bobsled (m = 100 kg) from rest to a speed of 6 m/s. How much kinetic energy does the bobsled gain?


b. How much work did the bobsled team do? Explain how you found the answer.


c. If the teammates pushed the bobsled a distance of 8 m, how much force did they apply?


d. Later, the bobsled is racing down the track at a speed of 20 m/s. With the team in it, the bobsled has a total mass of 325 kg. How much work must be done by friction to slow down the bobsled to 17 m/s?

Answer 36

KE=(1/2)mv^2/r

Answer 37

actually no r

Answer 38

that's the centripetal force equation, KE is just (1/2)mv^2

Answer 39

KE=(1/2)mv^2
KE=(1/2)(100)(6)^2
KE=1800J

Answer 40

good

for b) all of the work they did must have gone into increasing KE (assuming 0 friction) so it's also 1800J of work

Answer 41

for c) [I am assuming they are talking about the same work from b?) W = Fd, solve for F

Answer 42

1800=(f)8

Answer 43

good, solve for F

Answer 44

f=225N

Answer 45

good

for d) all the work goes into slowing the bobsled down so work = change in KE = ?

Answer 46

change is 3

Answer 47

because 20-17

Answer 48

getting there

since change in velocity is 3 what's the change in KE

Answer 49

would it be squared?

Answer 50

KE = (1/2)mv^2

change in KE = (1/2)m(deltav)^2 where deltav is your change in velocity

Answer 51

change in KE = (1/2)m(deltav)^2
change in KE = (1/2)(325)(17)^2

Answer 52

delta v is the ~change~ in velocity which we determined to be 20 - 17 = 3 m/s

Answer 53

change in KE = (1/2)m(deltav)^2
change in KE = (1/2)(325)(3)^2

Answer 54

good, then just evaluate all that

Answer 55

1462.5

Answer 56

J

Answer 57

good

Answer 58

a) displacement is net distance travelled, so you only need to consider the distance between the starting (A) and final (F)

Answer 59

4

Answer 60

must also include a direction since displacement is a vector

Answer 61

if the cyclist travels from A to F what direction are they going in? you can use the compass at the bottom left to give you a clue

Answer 62

East

Answer 63

good so displacement = 4 m east

Answer 64

for b) distance, you have to add up all the segments travelled

Answer 65

13

Answer 66

good (I just read the thing at the top that says each square is 10 m) so a is actually 40 m east and b) is 130 meters (no direction since displacement isn't a vector)

Answer 67

c) average speed = distance/time = ?

(make sure you are using distance not displacement)

Answer 68

130/30=4.3

Answer 69

good, units of speed are m/s as usual

Answer 70

for d) it's a similar calculation just using displacement instead of distance

displacement/time = ?

Answer 71

10/30=0.3

Answer 72

displacement is 40 not 10

Answer 73

40/30=1.3

Answer 74

good, since displacement and velocity are vectors we specify that it is 1.3 m/s east

Answer 75

e) acceleration = change in velocity/time = ?

Answer 76

3/2=1.5

Answer 77

good, units are m/s/s as usual

Answer 78

You stand on a bridge above a river and drop a rock into the water below from a height of 53 m. (Assume no air resistance.)

a. What is the acceleration of the rock as it falls?

b. How long does it take to hit the water?

c. How much more time would it take if instead of dropping the rock straight down, you threw it horizontally with a speed of 5 m/s? Explain.


d. You throw another rock at an angle of 30° above the horizontal with a speed of 12 m/s. What are the x- and y-components of the rock's speed?

e. How long does it take the rock to reach its maximum height?

Answer 79

a) initial PE = final KE

so mgh = (1/2)mv^2 solve for v

Answer 80

m(0)(53) = (1/2)mv^2

Answer 81

g is 9.8 not 0

Answer 82

it says no air resistance that is why i put zero

Answer 83

m(9.8)(53) = (1/2)mv^2

Answer 84

good, solve for v

Answer 85

32.23

Answer 86

good, units are m/s as usual

for b) you can use a kinematics equation

deltax = (1/2)at^2

where deltax is height and a is the acceleration due to gravity

Answer 87

deltax = (1/2)at^2
53=(1/2)(9.8)t^2

Answer 88

t=3.29

Answer 89

good

for c) throwing the rock horizontally has no effect on the y-components of velocity so the rock will take the same time to fall

Answer 90

d) x component is velocity * cos(theta), y component is velocity * sin(theta)

Answer 91

12*cos(30)=10.39
12*sin(30)=206

Answer 92

check that second one again, it looks suspect

Answer 93

12*cos(30)=10.39
12*sin(30)=6

Answer 94

awesome, units are m/s on both as usual

Answer 95

okay for e

Answer 96

kinematics equation 2 - since the velocity of rock is 0 at max height,

vf = v0 + at

where vf is 0, v0 is the initial y-component of velocity (make sure you are using the resolved y- component, including the angle), a is the acceleration due to gravity (use -9.8 here) and solve for t

Answer 97

vf = v0 + at
0=6+(-9.8)t

Answer 98

good, solve for t

Answer 99

t=0.61

Answer 100

good, units are seconds as usual

Answer 101

April is moving into a new apartment. She is moving a box of books that has a mass of 30 kg.

a. What is the weight of the box?

b. What is the normal force acting on the box when it sits on the floor? Explain how you found the answer.


c. If she pushes on the box with a force of 100 N and there is no friction, what is the acceleration of the box?

d. If the coefficient of static friction between the box and the floor is 0.3, what is the maximum force of static friction acting on the box?


e. How hard must she push on the box to get it to slide?


f. If she pushes on the box with a force of 100 N, and the coefficient of kinetic friction is 0.25, what is the net force on the box in the x-direction?




g. Based on your answer to part (f), what is the acceleration of the box?

Answer 102

a. 30*9.8=294

Answer 103

good, units of weight are N as usual

for b) since the box is lying flat, normal force = weight force = answer from a)

Answer 104

c) F = ma, solve for a

Answer 105

3994=30a
a=1331.3

Answer 106

where are you getting 3994

the force is given as 100N

Answer 107

100 = 30*a solve for a

Answer 108

3.3

Answer 109

m/s

Answer 110

good a = 3.3 m/s/s

d) F = μN, N is the normal force from part b), find F

Answer 111

μ is the coefficient of static friction

Answer 112

F=(0.3)294
F=88.2N

Answer 113

good

Answer 114

e) the force to start sliding is equal to the max static fric force so copy the answer from d)

Answer 115

f) net force = applied force - μN, make sure you are using the kinetic friction constant for μ

Answer 116

0.3*88.2
26.46

Answer 117

μ is 0.25 not 0.3 this time, different constant

Answer 118

0.25*88.2=22.05

Answer 119

good, but we are looking for net force not just friction force

so applied force = 100N

applied force - friction force = ?

Answer 120

100-22.05=77.95N

Answer 121

good

for g) set that force equal to ma and solve for a

Answer 122

f=ma
77.95=30a

Answer 123

2.6

Answer 124

good, 2.6 m/s/s

and that's it

Answer 125

A car of mass 1250 kg drives around a curve with a speed of 22 m/s.

a. If the radius of the curve is 35 m, what is the centripetal acceleration of the car?






b. The maximum friction that the car's tires can apply is 20,000 N. What is the radius of the smallest circle the car can drive in at a speed of 22 m/s?

Answer 126

a = v^2 / r = ?

Answer 127

22=v^2/35

Answer 128

we are solving for a not v here

v is already given as 22 m/s

Answer 129

a=22^2/35
13.86

Answer 130

I got something a little diff 13.83 m/s/s

Answer 131

anyway for b) set force equal to 20,000, v = 22 and solve for r where

F = mv^2/r

Answer 132

20000=1250(22)^2/r
r=121/4
r=30.25

Answer 133

awesome, r is a distance so meters are the units

Answer 134

Alan hits a hockey puck of mass 0.3 kg across a hockey rink with a speed of 23 m/s to the south.

a. What is the momentum of the hockey puck?





b. If the puck started at rest and the time of impact was 0.1 seconds, what force did Alan apply to the puck when he hit it?












c. The puck slides across the rink, losing 2 m/s of speed due to friction. What impulse is applied by friction?












d. When Alan's puck is moving at 21 m/s south, it collides with an ice skate of mass 2 kg sitting on the rink. Alan's puck bounces backward at a speed of 4 m/s. What is the momentum of the puck-skate system before the collision?








e. What is the speed of the ice skate after the collision?


A toy dart gun uses a spring with a spring constant of 35 N/m. To use the dart gun, you compress the spring by pushing in a dart of mass 0.002 kg.

a. If you compress the spring by 0.02 m, what force is the spring exerting on the dart?








b. With the spring compressed 0.02 m, how much elastic potential energy is stored in the spring?







c. If you release the spring, it pushes the dart forward. What is the kinetic energy of the dart when it reaches the natural length of the spring? Explain how you found the answer.








d. What is the speed of the dart when it reaches the natural length of the spring?

Answer 135

p=mv
p=0.3*23
p=6.9 kg m/s

Answer 136

good, that's a)

Answer 137

okay and for b it would be 0?

Answer 138

for b) set change in momentum equal to impulse and solve for F

Answer 139

remember that impulse = F * delta t

Answer 140

F = mΔv / Δt = 0.3kg * 23m/s / 0.1s = 69 N

Answer 141

well done

for c) simply calculate the change in momentum, that's your impulse

Answer 142

mpulse = mΔv = 0.3kg * -2m/s = -0.6 kg·m/s

Answer 143

awesome

for d) before the collision you only need to consider the momentum of the puck = ?

Answer 144

p = 0.3kg * 21m/s = 6.3 kg·m/s

Answer 145

and 6.3 kg·m/s = 0.3kg * -1.2m/s + 2kg * v
v = 6.7kg·m/s / 2kg = 3.3 m/s for e

Answer 146

the puck bounces back with velocity 4 m/s so just replace that -1.2 with -4

Answer 147

6.3 kg·m/s = 0.3kg * -4m/s + 2kg * v
v = 6.7kg·m/s / 2kg = 3.3 m/s

Answer 148

you need to re-do the v calculations since you replaced a value

Answer 149

(6.3) = (0.3)(-4) + 2* v solve for v

Answer 150

(6.3) = (0.3)(-4) + 2* v
v=3.75

Answer 151

awesome, units are m/s as usual

Answer 152

for the gun problem

a) F = -kx as usual (not sure about the sign convention here but I would presume a negative delta x since the spring is being compressed?)

Answer 153

F=-0.02x

Answer 154

0.02*35

Answer 155

F=0.7

Answer 156

good, units are N as usual

Answer 157

b) U = (1/2)kx^2 = ?

Answer 158

U = (1/2)kx^2 =
U=(1/2)(0.02)(35)^2

Answer 159

the 35 is k which is not the term that gets squared, it's the 0.02 m that gets squared

Answer 160

12.25

Answer 161

U = (1/2)kx^2 =
U=(1/2)(35)(0.02)^2

Answer 162

U=0.007

Answer 163

good

for c) all the PE from b) gets converted to KE so the answer is also 0.007J

Answer 164

for d) set that energy equal to (1/2)mv^2 and solve for v

Answer 165

0.007=(1/2)(0.002)(35)^2

Answer 166

v is not 35, we are solving for v

Answer 167

0.007=(1/2)(0.002)v^2

Answer 168

so like this

Answer 169

yes

Answer 170

2.65

Answer 171

awesome, units are m/s/s as usual

Answer 172

and that's it, gonna go hit the sack

Answer 173

Same with me thanks so much