A 480 G Peregrine Falcon
A 480 g peregrine falcon reaches a speed of 75 yard/s in a vertical dive chosen a stoop. (Figure 1)
If we assume that the falcon speeds up under the influence of gravity only, what is the minimum peak of the
dive needed to reach this speed?
The sledder shown in (Figure 1) starts from the pinnacle of a frictionless hill and slides down into the valley.
Suppose that h1 = 4.3 m and h2 = 5.5 m . What initial speed half-dozen does the sledder demand to only brand it over the
next hill?
A xix kg child slides downwards a 2.6-m-high playground slide. She starts from remainder, and her speed at the lesser is
2.seven grand/s. What isouthward the change in thermal energy of the slide and the seat of her pants?
A cyclist is benumbed at fourteen grand/south when she starts down a 460 k long gradient that is thirty m high. The cyclist and her
bicycle accept a combined mass of seventy kg. A steady xiv N drag strength due to air resistance acts on her equally she
coasts all the style to the bottom. What is her speed at the bottom of the gradient?
In a physics lab experiment, a spring clamped to the tabular array shoots a 22 g ball horizontally. When the jump
is compressed 22 cm , the ball travels horizontally 4.8 1000 and lands on the floor 1.5 m below the point at
which it left the spring. What is the spring constant?
A ii kg mass compresses a leap with bound constant 1210 N/grand past a distance 0.95 g . The spring is
released and launches the mass on to a frictionless floor. On the floor there is a 2 m long mat with
coefficient of friction 0.2. What is the final velocity of the mass afterward is passes the friction mat?
A 2.2 kg mass compresses a spring past a altitude 0.5 m . The mass is released from remainder. The mass
travels over equal patches of rough surfaces with coefficient of friction µk= 0.ii and length 0.ix k . If the
spring constant of the spring is 920 N/m , how many (including the decimal fraction) mats does information technology take to
finish the mass? The respond could be larger than 6.
HW
32
#
I
v2
=
ugh
.
29.5
one thousand-
_
v2
/
29
=
752/2-nine.eight=286.69-7
#
2
112m
v2
=
Mg
(
h
z
-
h
,
)
7
=
Zglhz
-
h
,
)
=
219.00
)
(
five.5
-
4.
3)
=4.84M
#
three
Mgh
=
112mFive
+
estrus
³
49
.
9.8.2.6
)
-412.19
¥
2.vii
2)
E-
rut
³
estrus
-
-414.8650
#
4
¥ 003
=
42mV
,
Ztmgh
=
Fx
+
112mV
>
2
.
460
(5)
(7011144+170119.81130)=(141146011-i.51170)
V22
27440=6440+3542
2103 ¥
-
-3542
v22
-
-6500
=
Z4.49
-
6440
-6440
21000
#
5
house
b-
=
2h19
Z(
1.
5)
19.eight
=
.
55
³
4.001.55=00.67
³
.
022100.67
/
¥
ZZF-34.MN/m@
#
6
112
KXZ
_
Mmg
D=
112m42
112
¥
1210
¥
(-9512-ane.2*2-ix.002)=one/ii
¥
ZVFZ
546
-38.42=42
Z3.2 1000
#
vii
PE
-
-
42192011.52
)
=
115J
³
kik
mg
¥
L
C.
2)
(
2.
2)
(nine-811.9)=3.800
115/3.800
=Z g
A 480 G Peregrine Falcon,
Source: https://www.studocu.com/en-us/document/texas-state-university/general-physics-i/phys-homework-exam-3/31383212
Posted by: milesupor1961.blogspot.com
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