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”)
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) 7.5 0.2
(c) 2.8 109
(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.
(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 ?
(a) (2.69 104) (6.281 109) / (5.37 104)
(b) (3.1416 102) (2.355 105) / (1.082 109)
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
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
Average speed = 52.94 km /h
total distance = 90 km
(a) Which boat wins? (Or is it a tie?)
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
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) How long does the race take?
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]
distance travelled = average velocity*time
= 0.5(33.1 + 25)*3.2
= 92.96 m
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?
(b) How far does the car travel in the 3.20-s time period?
Thanks for the help
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
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
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?
delta t 2* 0.4/(82.4+16.4) hr * 3600s/hr=29.15 s
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
Whta is the direction (in degrees S of E)? Answer in units ofo
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
4/sin(26.6) = 8.9 blocks
Read more: http://wiki.answers.com/Q/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#ixzz1u3eEjhBO
Read more: http://wiki.answers.com/Q/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#ixzz1u3eDYM00
F1=120cos60i+120sin60j and F2=80cos75i+80sin75j
?=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
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
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.
– 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
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
speed = ?(2ad),
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
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.
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
Vbg = v[ (Vbw)^2 -(Vwg)^2 ]
= v(15m/s)^2 – (1m/s)^2
The angle through which the boat drift is
? = sin^-1 (1 m/s /15m/s )
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
(a) What is the velocity of the boat relative to shore?
° (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?
River has a steady speed of ,vs
Width of the river,d
Student can swim at a speed of, v
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,
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,
time required for total trip
the trip in still water takes, tb=2d/v
It is clear that,
No, it happens only in the direction against to ‘
(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.
(b) Using the same variables, how much time tdown does it takes to swim back downstream to the starting point?
(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
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.)
(e) Which is larger, ta or tb?
Is it always larger?
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
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)
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?
? radians below horizontal
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..
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
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.
= ? 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.
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
Please Rate it………….Thanks
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
answer in N
define positive as downward, we have
W – Fdrag = ma
mg – kv2 = ma
at terminal speed, a = 0
mg = kv2
k = mg/v2
k = (76.5kg)(9.8m/s2)/(58.7m/s)2
k = 0.218 kg/m
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) 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?
Please help!! Thank You
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 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.
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
Conservation of momentum
Eqs of constant acceleration with time eliminated
If the contact between ball and ground lasted 2.4 ms, what average force was exerted on the ball?
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
It goes north east a bit.
tan alpha = 500/270
alpha = 61 degrees
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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?