Physics study

dimensions of B = dimension of x/(dimension of t)2 = L/T2 or LT-2

“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”)

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

2,4,2,2

3.00e+08
2.9979e+08
2.997925e+08

9.90e+02
6.7
19.9

3.14e+09
0.06838

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

time= distance/speed= 2/100 =0.02 seconds

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

Boat A, 52, 0

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

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.

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

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

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

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

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 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

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

Give your information
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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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

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
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

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

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

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

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

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)

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

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.

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

Please Rate it………….Thanks

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)

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) 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]
3 years ago Report Abuse

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

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 mass is not important in an acceleration problem
use:
a=gsin?
a=9.8sin4
a=0.684m/ss