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dimensions of B = dimension of x/(dimension of t)2 = L/T2 or LT-2
(a) Suppose that the displacement of an object is related to time according to the expression x = Bt2. What are the dimensions of B?
“x” is again “displacement”, so replace it with “L”:
L = A sin(2?ft).

They tell you that “sin(2?ft)” is dimensionless. You can throw out any dimensionless number that multiples another number. So throw it out:
L = A

Now it’s simple: With what should you replace “A” in order to make the left side equal the right side? (Hint: “L”)

b) A displacement is related to time as x = A sin(2?ft), where A and f are constants. Find the dimensions of A. [Hint: A trigonometric function appearing in an equation must be dimensionless.]
The smaller one of the two striped bass measurer A length of 95.42cm
The larger fish measures 120.6 cm
Hence the total length of fish caught for the day will be = 95.42 + 120.6 cm
Calculating : total length = 216.02 cm
Hence , rounded for 4 significant digits

Ans : 216.0 cm

A fisherman catches two striped bass. The smaller of the two has a measured length of 95.42 cm (two decimal places, four significant figures), and the larger fish has a measured length of 120.6 cm (one decimal place, four significant figures). What is the total length of fish caught for the day? (Enter your answer to the correct number of digits.)
2,4,2,2
• How many significant figures are there in each of the following?
(a) 7.5 0.2
3.690 0.004
(c) 2.8 109
(d) 0.024
3.00e+08
2.9979e+08
2.997925e+08
The speed of light is now defined to be 2.997 924 58 108 m/s.
(a) Express the speed of light to three significant figures.
(b) Express the speed of light to five significant figures.
Express the speed of light to seven significant figures.
9.90e+02
6.7
19.9
Carry out the following arithmetic operations. (Enter your answer to the correct number of significant figures.)
(a) the sum of the measured values 944, 36.7, 0.97, and 8.5
(b) the product 2.9 2.323
(c) the product 6.33 ?
3.14e+09
0.06838
Using your calculator, find, in scientific notation with appropriate rounding, the value of each of the following.
(a) (2.69 104) (6.281 109) / (5.37 104)
(b) (3.1416 102) (2.355 105) / (1.082 109)
We know the conversion between mile and feet is
1 mile = 5280 ft
And it is given that 1 fathom = 6ft
1 ft = (1/6)fathom
Then the distance between Earth to Moon = 240,000 miles
= (240,000 miles)(5280ft/mile)
= 1267200000 ft
= (1267200000)(1/6)fathom
= 211200000fathom
A fathom is a unit of length, usually reserved for measuring the depth of water. A fathom is approximately 6ft in length. Take the distance from Earth to the Moon to be 240,000 miles, and use the given approximation to find the distance in fathoms.
time= distance/speed= 2/100 =0.02 seconds
The speed of a nerve impulse in the human body is about 100 m/s. If you accidentally stub you toe in the dark, estimate the time it takes the nerve impulse to travel to your brain. (Assume that you are approximately 2.0 m tall and that the nerve impulse travels at uniform speed.)
Average speed = total distance / total time.

distance = speed* time

d1 = 80km /h * 30 min. = 80 * 30 /60 = 40 km

d2 = 100 *12 / 60 = 20 km

d3 = 40 *45 /60 = 30 km

( time is divided by 60 to convert into hour)

total distance d = 40+ 20 +30 = 90 km

total time = time of traveling + time at rest = (30 + 12 + 45) +15

t = 102min. = 102 /60 = 1.70 hr.

s = d /t = 90 /1.70 = 52.94 km/h

a)
Average speed = 52.94 km /h

b)
total distance = 90 km

A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 30.0 min at 80.0 km/h, 12.0 min at 100 km/h, and 45.0 min at 40.0 km/h and spends 15.0 min eating lunch and buying gas. (a) Determine the average speed for the trip. (b) Determine the distance between the initial and final cities along the route.
Boat A, 52, 0
Two boats start together and race across a 52 km wide lake and back. Boat A goes across at 52 km/h and returns at 52 km/h. Boat B goes across at 26 km/h, and its crew, realizing how far behind it is getting, returns at 78 km/h. Turnaround times are negligible, and the boat that completes the round-trip first wins.
(a) Which boat wins? (Or is it a tie?)
Average velocity = displacement/time
Velocity is vector and speed is scalar
Taking +x axis as +ve direction, while -xe axis as -ve direction

a) In ther first half , average velocity= L/t1 m/s

b)In the second half, average velocity = -L/t2 m/s

c) In the round trip, average velocity = 0/(t1+t2)=0m/s ( as the displacement is 0)

d) Average speed = total distance/total time= ( L+L)/(t1 +t2)= 2L/(t1 +t2) m/s

An athlete swims a length L of a pool in a time t1 and makes the return trip to the starting position in a time t2. If she is swimming initially in the positive x-direction, determine the average velocity symbolically in the first half of the swim.
Speed of Tortoise = 0.13m/s = S

Speed of Hare = 20x.13 = 2.6m/s = S’

let time taken by tortoise to reach final line = t sec

time taken by hare till tortoise wins = t+(5×60) = t-300 [it takes rest for 300 s where distance travelled is 0]

Distance travelled by tortoise = 0.13t

Distance travelled by hare = 2.6(t-300)

Difference between the distances of tortoise and hare is 0.35 m.

0.13t – 2.6(t-300) = 0.35

=> t = 315.65 s

Distance of the race = Time taken by tortoise x Speed of tortoise = 315.65 x 0.13 = 41.0342 m

The race is 41.0342 m long.

A tortoise can run with a speed of 0.13 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 5.0 minutes. The tortoise wins by a shell (35 cm).
(a) How long does the race take?
qa
v = u + at

25 = 33.1 + 3.2a
3.2a = – 8.1
a = – 8.1/3.2
= – 2.53 m/s^2 [ the minus sign indicates direction opposite to that of motion]
———————

qb
distance travelled = average velocity*time
= 0.5(33.1 + 25)*3.2
= 92.96 m

Car traveling east at 33.1 m/s passes a trooper hiding at the roadside…?
A car traveling east at 33.1 m/s passes a trooper hiding at the roadside. The driver uniformly reduces his speed to 25.0 m/s in 3.20 s.

(a) What is the magnitude and direction of the car’s acceleration as it slows down?
magnitude ____m/s2

(b) How far does the car travel in the 3.20-s time period?
_______meters

Thanks for the help

since the brakes were applied, it would be a uniform accelerationof -1 m/s2, right?

so the first thing is that the train won’t be moving for the full40 seconds. the brakes will stop the train after only 20seconds
v=v0+at
0=20-1.t
t=20s

x=1/2 at^2+v0t
0.5*-1*20^2+(20)(20)
x=200m

Anonymous asked A train is traveling down a straight track at 20 m/s when the engineer applies the brakes, resulting in a uniform acceleration of 1 m/s2. how far will the train move in a 40s time interval, starting at the instant the brakes are applied? I need the steps broken down if that is at all possible
a) velocity is given by v(t)=v0+at, here v0=0, a=a1 and t=t1, so we have that

v(t)=a1 t1

b) distance traveled = v0t + 1/2 at^2 so here, that means

dist = 1/2 a1 t1^2

c) the distance the car travels once the brakes are applied is

dist = v0 t + 1/2 at^2

now, v0=a1 t1 and a=-a2

t=t2 so you can write

dist in the braking phase = v0 t1 t2 -1/2 a2 t2 ^2

now, the problem does not state whether the car brakes to rest, if it does we can write t2 in terms of t1

v(t)=v0 + at

in this phase, if the final velocity at t=t2 is zero, and v0=a1 t1, we have that:

0=a1 t1 – a2 t2 or t2= a1 t1/a2

and you can use this expression in your final answer for part c

A car starts from rest and travels for t1 seconds with a uniform acceleration a1.?
The driver then applies the brakes, causing a uniform acceleration a2. If the brakes are applied for t2 seconds, determine the following. Answers are in terms of the variables a1, a2, t1, and t2.

(a) How fast is the car going just before the beginning of the braking period?

(b) How far does the car go before the driver begins to brake?

(c) Using the answers to parts (a) and (b) as the initial velocity and position for the motion of the car during braking, what total distance does the car travel?

Thank you!

deltax=400m=0.4 km
vaverage=x/t
0.5(vi+vf)=delta x/t
delta t 2* 0.4/(82.4+16.4) hr * 3600s/hr=29.15 s
A train 400m long is moving on a straight track with a speedof 82.4 km/h. The engineer applies the brakes at a crossing,and later the last car passes the crossing with a speed of 16.4km/h. Assuming constant acceleration, determine how long thetrain blocked the crossing. Disregard the width of thecrossing.
The first putt = 7.8m i
the second putt = – 6.89 j
the displacement directly to hole
R = ? ( 7.8)2 + (6.89)2 =10.4m
Direction ? = tan-1( -6.89 / 7.8)= – 41.45 0
A golfer takes two putts to get his ball into the hole once he ison the green. The first putt displaces the ball 7.8 m east, and thesecond 6.89 m south. What displacement would have been needed toget the ball into the hole of the first putt? Answer in units ofm.

Whta is the direction (in degrees S of E)? Answer in units ofo

A girl delivering newspapers covers her route by traveling 10.00 blocks west, 4.00 blocks north, and then 2.00 blocks east

Opposite = 4 (related to y axis on 4 blocks north)
Adjacent = 8 (related to x axis on 10 b west – 2 blocks east)
tan -1 x (4/8)
tan theta – 26.6 degrees
since you are looking for the hypotenuse, then sin theta = opp/hyp
sin 26.6=4/hypotenuse
4/sin(26.6) = 8.9 blocks

A girl delivering newspapers covers her route by traveling 10.00 blocks west 4.00 blocks north and then 2.00 blocks east. What is her resultant displacement?

F1=120cos60i+120sin60j and F2=80cos75i+80sin75j
R=F1+F2
R11=(120cos60+80cos75)i+(120sin60+80sin75)j=30.706i+181.197j
(a)Magnitude R1=183.78
?=tan-1(181.97/30.706)=80.38 above the +xaxis
(b)To have net force you would exert the same magnitude offorce but in the opposite direction. So
R2=183.78
?=180+80.38=260.68
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The helicopter view in the figure below shows two people pulling ona stubborn mule. Assume that F1 is 120 N, and F2 is 80 N. The forces are measured in units of newtons(N).
This Picture is not showing up but Force 1 isat an angle of 60.0 and force 2 is at 75
(a) Find the single force that is equivalent tothe two forces shown.
(b) Find the force that a third person would have to exert on the mule to make the net force equal to zero
Answered by Anonymous 29 minutes later Rating:4 Stars Given that
angle ? = 25o
initial speed uo = 10 m/s
time of fight t = 2.90 s
—————————————————
Let x = height of the building
When the brick hits the ground its displacement is – x.
Hence,
– x = ut) + (1/2)(g)(t^2)
– x = uo(sin 25)(2.90) + (1/2)(-9.8)(t^2)
– x = (10)(sin 25)(2.90) – 4.9(2.90)^2
x = 28.953 m

So the height of building is = 28.953m

Anonymous asked A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and with an initial speed of 10 m/s. If the brick is in flight for 2.9 s, how tall is the building?
Here is how you shoul approach this problem.

Normally you are supposed to find accleration of the car as
a = g sin(?),
but they already gave you the value of a:
a = 2.43 m/s²

Use fomula for speed of accelerated motion
where d is distance travelled by the car down the slope:
d = 45m.

Now you need horizontal and vertical components of velocity:
Vx = speed * cos(?)
Vy = speed * sin(?)

Write equation of motion of car in free fall:
y = 40m – Vy t – gt²/2

Equate y to zero, find t. Now you can find the distance from the cliff x:
x = Vx t

car is parked on a cliff overlooking the ocean ……..?
A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 15.0° below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 2.43 m/s2 for a distance of 45.0 m to the edge of the cliff, which is 40.0 m above the ocean.

Find the car’s position relative to the base of the cliff when the car lands in the ocean.

Find the length of time the car is in the air.

I got the answer is 3.39 m/s

Yf = Yi + Vit + .5at^2
0 = 40 + (0)t + .5(-9.8)t^2
t = 2.857

2.98 – 2.86 = .12

.12 * 343 = 41.16

41.16 = ?(x^2 + 40^2)

x = 9.70

Vxi = (9.70)/(2.86)

Vxi = 3.39

A soccer player kicks a rock horizontally off a 40.0 m high cliffinto a pool of water. If the player hears the sound of the splash2.98 s later, what was the initialspeed given to the rock? Assume the speed of sound in air to be 343m/s.
The velocity of the boat with respect to the ground is

Vbg = v[ (Vbw)^2 -(Vwg)^2 ]

= v(15m/s)^2 – (1m/s)^2

= 14.966m/s

The angle through which the boat drift is

? = sin^-1 (1 m/s /15m/s )

= 3.822o

the angle made with boat with the direction of the river is

f = 90- 3.822 = 86.17 north of east.

b) If the river is 200m the drift in distance is

tan 3.822 = x / 200m

x = 200 tan 3.822

= 13.36m

A river flows due east at 1.00 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 15.0 m/s due north relative to the water.
(a) What is the velocity of the boat relative to shore?
m/s
° (north of east)

(b) If the river is 200 m wide, how far downstream has the boat moved by the time it reaches the north shore?
m

SOLUTION:
River has a steady speed of ,vs
Width of the river,d
Student can swim at a speed of, v
(a)
In order to swim up stream, the student must swim against the river’s
current. so relative speedof the student must be , v-vs
thus, requireed time is,
tup= d/v-vs
_________________________________________________________________
_________________________________________________________________
(b)
In order to swim downstream, the student must swim in the river’s
current direction. so relative speedof the student must be , v+vs
thus, requireed time is,
tup= d/v+vs
_________________________________________________________________
_________________________________________________________________
(c)
time required for total trip
ta=(d/v-vs) +(d/v+vs)
=2dv/v2-vs2
____________________________________________________
____________________________________________________
(d)
the trip in still water takes, tb=2d/v
___________________________________________________
___________________________________________________
(e)
It is clear that,
2d/v >2dv/v2-vs2
Hence,
tb>ta
No, it happens only in the direction against to ‘
river flow
A river has a steady speed of vs. A student swims upstream a distance d and back to the starting point.
(a) If the student can swim at a speed of v in still water, how much time tup does it take the student to swim upstream a distance d? Express the answer in terms of d, v, and vs.
tup =

(b) Using the same variables, how much time tdown does it takes to swim back downstream to the starting point?
tdown =

(c) Sum the answers found in parts (a) and (b) and show that the time ta required for the whole trip can be written as
ta =
2d/v
1 – vs2/v2
.
(Do this on paper. Your instructor may ask you to turn in this work.)
(d) How much time tb does the trip take in still water? (Use the following as necessary: d and v.)
tb =

(e) Which is larger, ta or tb?
ta
tb

Is it always larger?
Yes
No

let’s make x axis pointing east, y axis pointing north

Vx = 300 + 105 cos36 = 384.95

Vy = 105 sin36 = 61.72

V = sqrt(Vx2 + Vy2 ) = 390 mi/h

? = tan-1 (Vy/Vx) = tan-1(61.72/384.95) = 9.1 degree

new velocity is 390 mi/h in a direction 9.1 degrees north of east

A jet airliner moving initially at 300 mi/h due east enters a region where the wind is blowing at 105 mi/h in a direction 36.0 degrees north of east. What is the new velocity of the aircraft relative to the ground?
Best Answer – Chosen by Voters
h = 0.5 g t^2

t = sqrt (2h / g)

The speed of the mug when it left the counter

Vx = d / sqrt (2h / g)

The vertical component of this velocity is:

Vy = g sqrt (2h / g) = sqrt (2 g h)

The angle will be given by:

InverseTan (Vy / Vx)

Projectile motion: In a local bar, a customer slides an empty beer mug down the counter for a refill.?
The bartender is momentarily distracted and does not see the mug, which slides off the counter and strikes the floor at distance d from the base of the counter. The height of the counter is h.

(a) With what speed did the mug leave the counter?

(b) What was the direction of the mug’s velocity just before it hit the floor?

i suck at these problems, i can never think of the variables as jus being a constant…and the program where i have to submit the answer always says syntax is wrong? i need to see if my answer is wrong or im just inputting it incorrectly..

A Foot ball accelerates rest from to speed is v = 10 m/s
Initial speed is 0 m/s
total time is 0.16 s
Acceleration a = ?v / ?t
= 10 m/s – 0 m/s / 0.16s – 0s
= 62.5 m/s2
According toNewtons second law F = ma
= 0.50 kg * 62.5 m/s2
= 31.25 N
football punter accelerates a football from rest to a speed of10 m/s during the time in which histoe is in contact with the ball (about 0.16 s). If the football has a mass of 0.50 kg,what average force does the punter exert on the ball?
1
Weight = mass x acceleration
Earths acceleration is 9.8 m/s*2
1 kg = 2.2 lbs, so 2.0 lbs x 1 kg/2.2 lbs = 0.91 kg
The bag would have a weight of 9.8 x 0.91 = 8.9 N

1. 8.9 x 1/6 = 1.5 N

2. 8.9 x 2.64 = 23.5 N

The mass of the bag at all three locations is 0.91 kg. Mass does not change, the different locations only change its weight.

A bag of sugar weighs 3.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixth that on Earth?
= ? N
Repeat for Jupiter, where g is 2.64 times that on Earth.
= ? N
Find the mass of the bag of sugar in kilograms at each of the three locations.
Earth= kg
Moon= kg
Jupiter= kg
U need to find acceleration:

use:

delta x = 1/2 (v initial + v final)t
.85 m = 1/2 (340) t
.85 m = 170 t
0.005sec = t

Then to find the acceleration

v=v initial + at
340 = a(.005)
68000m/s^2 = a

Fnet = ma = (.0059kg)(68000m/s^2) = 401.2N

A 5.9 g bullet leaves the muzzle of arifle with a speed of 340 m/s. Whatforce (assumed constant) is exerted on the bullet while it istraveling down the 0.85 m long barrelof the rifle?
Sincethe salmon is 1.5 m long, 2/3 of its length is 1 m. This is thedistance travelled during the acceleration from 3 m/s to 6.9m/s.

The kinetic energy E of a body of mass m moving at velocity vis
E = 1/2 * m * v^2

So the work W necessary to make the difference in kinetic energy ofa salmon at 3 m/s and 6.9 m/s is
W = 1/2 * 64 kg * (6.9 m/s)^2 – 1/2 * 69 kg * (3 m/s)^2= 1235.52 J

Now work is force times distance
W = F * d

which you can solve for F
F = W / d

and inserting the distance (2/3 of the salmons length, seeabove):
F = 1235.52 J / 1 m = 1235.52 N

Nowthis is the net force necessary for the acceleration. This netforce is the sum of the tail fin force Ff and gravity Fg.
F = Ff + Fg
(the force of gravity acts downwards, so it will assume a negativevalue)

Solving this for the tail fin force gives
Ff = F – Fg

Now gravity on a salmon of mass m = 61 kg is
Fg = – g * m = – 9.81 m/s^2 * 64 kg = – 627.2 N

So the tail fin force is
Ff = 1235.52 N – (- 627.2 N) = 1862.72N

A Chinook salmon has a maximum underwater speed of 3.0 m/s,and can jump out of the water vertically with a speed of6.9 m/s. A record salmon has a lengthof 1.5 m and a mass of 63 kg. Whenswimming upward at constant speed, and neglecting buoyancy, thefish experiences three forces: an upward force F exertedby the tail fin, the downward drag force of the water, and thedownward force of gravity. As the fish leaves the surface of thewater, however, it experiences a net upward force causing it toaccelerate from 3.0 m/s to 6.9 m/s.Assuming the drag force disappears as soon as the head of the fishbreaks the surface and that F is exerted until two-thirdsof the fish’s length has left the water, determine the magnitude ofF.

a)

define positive as downward, we have
W – Fdrag = ma
mg – kv2 = ma

at terminal speed, a = 0

so

mg = kv2

k = mg/v2

k = (76.5kg)(9.8m/s2)/(58.7m/s)2

k = 0.218 kg/m

b)

use

mg – kv2 = ma

a = g – kv2/m

a = 9.8m/s2 – (0.218kg/m)(0.5(58.7m/s))2/76.5kg

a = 2.45 m/s2

A 76.5-kg skydiver reaches a terminal speed of 58.7 m/s with her parachute undeployed. Suppose the drag force acting on her is proportional to the speed squared, or Fdrag = kv^2.

(a) What is the constant of proportionality k? (Assume the gravitational acceleration is 9.8 m/s2.)

(b) What was the magnitude of the acceleration when she was falling at half terminal speed?

a) t = ?[2h/g]

b) a(x) = F/m

c) Xh = ½*a(x)*t² = ½*(F/m)*2gh = Fgh/m

d) a(x) = F/m; a(y) = g;
….a = ?(a(x)²+a(y)²)

a = ?[(F/m)² + g²]
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An object of mass m is dropped from the roof of a building of height h. While the object is falling, a wind bl?
An object of mass m is dropped from the roof of a building of height h. While the object is falling, a wind blowing parallel to the face of the building exerts a constant force F on the object.
(a) How long does it take the object to strike the ground? Express the time t in terms of g and h.

(b) Find an expression in terms of m and F for the acceleration ax of the object in the horizontal direction (taken as the positive x-direction).

(c) How far is the object displaced horizontally before hitting the ground? Answer in terms of m, F, g, and h.

(c) Find the magnitude of the object’s acceleration while it is falling, using the variables F, m, and g.

V1 = ?(2gy1) = -23.84 m/s
V2 = ?(2gy2) = 19.30 m/s

dP = m*dV = m*(V2 – V1) = 23.30 kg?m/s

Fav = dP/t = 23.30/.0024 = 9708 N up
Source(s):
Conservation of momentum
Eqs of constant acceleration with time eliminated

After falling from rest from a height of 29 m, a 0.54 kg ball rebounds upward, reaching a height of 19 m.?
If the contact between ball and ground lasted 2.4 ms, what average force was exerted on the ball?
magnitude:

The two forces are at right angles so Pythagorean theorem can be applied.
F^2 = 500^2 + 240^2
F = 554N

F = ma
554 = 270a
a = 2.05 m/s^2

Direction:

It goes north east a bit.

tan alpha = 500/270
alpha = 61 degrees
3 years ago Report Abuse

The force exerted by the wind on the sails of a sailboat is 500 N north…?
The force exerted by the wind on the sails of a sailboat is 500 N north. The water exerts a force of 240 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration?
The mass is not important in an acceleration problem
use:
a=gsin?
a=9.8sin4
a=0.684m/ss
A 3.0 kg object hangs at one end of a rope that is attached to a support on a railroad car. When the car accelerates to the right, the rope makes an angle of 4.0 degree with the vertical. Find the acceleration of the car.

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