PHY 1114 College Physics I Fall 2012 Professor Kenny L. Tapp HOMEWORK #1 SOLUTIONS
1. The aorta of a patient has a cross-sectional area of 2.1 cm2. Convert this area to m2.
Knowing that 100cm = 1m, we can set up the conversion factor…
2.1 cm2 = 2.1 cm•cm • 1m/100cm
0.021 cm • m • 1m/100cm = 0.00021 m2
2. Convert 20 m/s to mi/h.
Using the conversions 1609.34 m = 1 mile, 3600 s = 1 hour…
20 m/s • 1 mile/1609.34m • 3600 s/1hour
= 44.74 miles/hour
3. Use the Pythagorean Theorem to find the missing side of each right triangle.
a)
2
2
b)
2
a2 + b2 = c2
c = √( a2 + b2)
c = √( 6m2 + 10m2)
c = 11.6m
a +b =c
a = √( C2 - b2)
a = √( 12m2 – 10m2)
a = 6.6m
4. Determine the angle θ in the right triangle shown.
SOH-CAH-TOA
tanØ =0.5m/0.7m
Ø =tan-1(0.5/0.7)
Ø =35.5°
1 5. A monkey is chained to a stake in the ground. The stake is 3.00 m from a vertical
pole, and the chain is 3.40 m long. How high can the monkey climb up the pole?
x = 3.00 m, hypotenuse = C = 3.40m, y = ?
Use Pythagorean theorem x2 + y2 = C2 to solve for y
y 2 = C2 - x2
y = √( C2 - x2) = √( 3.4m2 – 3.0m2)
y = 1.60 m
6. Find the x and y components of this vector: F = 35 N at a direction of 65 degrees
with respect to the x-axis.
Fx = FcosØ = (35N)cos(65°) = 32N = Fx
Fy = FsinØ =(35N)sin(65°) = 15N = Fy
7. Graphically solve each expression using the vectors in the sketch.
a)
€
d)
€
 
A+C
 
2A + C
b)
€
e)
 
A −C


Ax + Cx
€
 
C−A
f)


Ay + Cy
€
€
2 c)
8.
A meteoroid is speeding through the atmosphere, traveling east at 18.3
km/s while descending at a rate of 11.5 km/s. What is its speed, in km/s?
The meteoroid’s speed is the magnitude of its velocity vector, here described in terms of
two perpendicular components, one directed toward the east and one directed vertically
downward. Let east be the +x direction, and up be the +y direction. Then the
components of the meteoroid’s velocity are Vx = +18.3 km/s and Vy = −11.5 km/s. The
meteoroid’s speed v is related to these components by the Pythagorean theorem.
From the Pythagorean theorem,
It’s important to note that the negative sign for Vy becomes a positive sign when this
quantity is squared. Forgetting this fact would yield a value for V that is smaller than Vx,
but the magnitude of a vector cannot be smaller than either of its components.
9.
In a mall, a shopper rides up an escalator between floors. At the top of the
escalator, the shopper turns right and walks 9.00 m to a store. The magnitude of
the shopper’s displacement from the bottom of the escalator to the store is 16.0
m. The vertical distance between the floors is 6.00 m. At what angle is the
escalator inclined above the horizontal?
Consider first the shopper’s ride up the escalator. Let the diagonal length of the
escalator be L, the height of the upper floor be H,
and the angle that the escalator makes with respect
to the horizontal be θ (see the diagram). Because L
is the hypotenuse of the right triangle and H is
L H opposite the angle θ, the three quantities are related
θ
by the inverse sine function:
L Up the escalator s D Entire view Now consider the entire trip from the bottom to the
top of the escalator (a distance L), and then from the
top of the escalator to the store entrance (a distance s). The right turn between these
two parts of the trip means that they are perpendicular (see the diagram). The shopper’s
total displacement has a magnitude D, and this serves as the hypotenuse of a right
triangle with L and s. Use the Pythagorean theorem to relate the three sides.
Solving
for the length L of the escalator gives
this result and the equation above to obtain the angle θ:
3 . We now use
10.
The punter on a football team tries to kick a football so that it stays in the air for a
long “hang time.” If the ball is kicked with an initial velocity of 25.0 m/s at an
angle of 60.0° above the ground, what is the hang time?
The vertical motion consists of the ball rising for a time t stopping and returning to the
ground in another time t. For the upward portion
Note: Vy = 0 m/s since the ball stops at the top.
V0y = v0 sin θ0 = (25.0 m/s) sin 60.0° = 21.7 m/s t = (21.7 m/s)/(9.80 m/s2) = 2.21 s The required "hang time" is 2t =
11.
.
A spelunker (cave explorer) drops a stone from rest into a hole. The speed of
sound is 343 m/s in air, and the sound of the stone striking the bottom is heard
1.50 s after the stone is dropped. How deep is the hole?
The stone requires a time, t1, to reach the bottom of the hole, a distance y below the
ground. Assuming downward to be the positive direction, the variables are related by
(with V0 = 0 m/s):
The sound travels the distance y from the bottom to the top of the hole in a time t2.
Since the sound does not experience any acceleration, the variables y and t2 are related
by (with a = 0 m/s2 and vsound denoting the speed of sound):
Equating the right hand sides of the above equations and using the fact that the total
elapsed time is t = t1 + t2, we have
Rearranging gives
Substituting values and suppressing units for brevity, we obtain the following quadratic
equation for t1: 4 From the quadratic formula, we obtain
The negative time corresponds to a nonphysical result and is rejected. The depth of the
hole is then found using the value of t1 obtained above:
12.
A golfer hits a shot to a green that is elevated 3.0 m above the point where the
ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40.0°
above the horizontal. It rises to its maximum height and then falls down to the
green. Ignoring air resistance, find the speed of the ball just before it lands.
The magnitude (or speed) v of the ball’s velocity is related to its x and y velocity
components (Vx and Vy ) by the Pythagorean theorem:
The horizontal component vx of the ball’s velocity never changes during the flight, since,
in the absence of air resistance, there is no acceleration in the x direction (ax = 0 m/s2).
Thus, Vx is equal to the horizontal component V0x of the initial velocity, or
.
Since V0 is known, Vx can be determined. The vertical component Vy of the ball’s
velocity does change during the flight.
The relation
may be used to find Vy2, since ay, y, and V0y are known
(V0y = V0 sin 40.0º).
The speed v of the golf ball just before it lands is
5 13.
A fire hose ejects a stream of water at an angle of 35.0° above the horizontal.
The water leaves the nozzle with a speed of 25.0 m/s. Assuming that the water
behaves like a projectile, how far from a building should the fire hose be located
to hit the highest possible fire?
The water exhibits projectile motion. The x component of the motion has zero
acceleration while the y component is subject to the acceleration due to gravity. In order
to reach the highest possible fire, the displacement of the hose from the building is x,
where (recall ax = 0 m/s2),
with t equal to the time required for the water the reach its maximum vertical
displacement. The time t can be found by considering the vertical motion.
When the water has reached its maximum vertical displacement, Vy = 0 m/s. Taking up
and to the right as the positive directions, we find that
and
Therefore, we have
14.
You step onto a hot beach with your bare feet. A nerve impulse, generated in
your foot, travels through your nervous system at an average speed of 110 m/s.
How much time does it take for the impulse, which travels a distance of 1.8 m, to
reach your brain?
Since the average speed of the impulse is equal to the distance it travels divided by the
elapsed time, the elapsed time is just the distance divided by the average speed.
The time it takes for the impulse to travel from the foot to the brain is
6 15.
The left ventricle of the heart accelerates blood from rest to a velocity of 26 cm/s.
(a) If the displacement of the blood during the acceleration is 2.0 cm, determine
its acceleration (in cm/s2). (b) How much time does blood take to reach its final
velocity?
We know the initial and final velocities of the blood, as well as its displacement.
Therefore,
can be used to find the acceleration of the blood. The time it
takes for the blood to reach it final velocity can be found by using
.
a. The acceleration of the blood is
b. The time it takes for the blood, starting from 0 cm/s, to reach a final velocity of +26
cm/s is
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