The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.
All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.
If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.
Answer:
The horizontal component of the velocity is 188 m/s
The vertical component of the velocity is 50 m/s.
Explanation:
Hi there!
Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:
We can find vx using the following trigonometric rule of a right triangle:
cos α = adjacent / hypotenuse
cos 15° = vx / 195 m/s
195 m/s · cos 15° = vx
vx = 188 m/s
The horizontal component of the velocity is 188 m/s
To calculate the y-component we will use the following trigonometric rule:
sin α = opposite / hypotenuse
sin 15° = vy / 195 m/s
195 m/s · sin 15° = vy
vy = 50 m/s
The vertical component of the velocity is 50 m/s.
Answer:
Te ayudo con una de prueba $
Explanation:
Answer:
C) 1.0 m
Explanation:
The component of the velocity parallel to the sidewalk is:
vₓ = v cos θ
vₓ = 0.1 m/s cos 45°
vₓ = 0.0707 m/s
The distance traveled after 14 seconds is:
d = vₓ t
d = (0.0707 m/s) (14 s)
d = 0.99 m
Closest answer is C) 1.0 m.
Answer:
solution:
to find the speed of a jogger use the following relation:
V
=
d
x
/d
t
=
7.5
×m
i
/
h
r
...........................(
1
)
in Above equation in x and t. Separating the variables and integrating,
∫
d
x
/7.5
×=
∫
d
t
+
C
or
−
4.7619
=
t
+
C
Here C =constant of integration.
x
=
0 at t
=
0
, we get: C
=
−
4.7619
now we have the relation to find the position and time for the jogger as:
−
4.7619 =
t
−
4.7619
.
.
.
.
.
.
.
.
.
(
2
)
Here
x is measured in miles and t in hours.
(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),
to get:
= −
4.7619
=
1
−
4.7619
= −
3.7619
or x
=
7.15
m
i
l
e
s
(b) To find the jogger's acceleration in m
i
l
/
differentiate
equation (1) with respect to time.
we have to eliminate x from the equation (1) using equation (2).
Eliminating x we get:
v
=
7.5×
Now differentiating above equation w.r.t time we get:
a
=
d
v/
d
t
=
−
0.675
/
At
t
=
0
the joggers acceleration is :
a
=
−
0.675
m
i
l
/
=
−
4.34
×
f
t
/
(c) required time for the jogger to run 6 miles is obtained by setting
x
=
6 in equation (2). We get:
−
4.7619
(
1
−
(
0.04
×
6 )
)^
7
/
10=
t
−
4.7619
or
t
=
0.832
h
r
s
