Solved Quiz 3 for Engineering Physics I | PHY 303K, Quizzes of Physics

Material Type: Quiz; Professor: Antoniewicz; Class: ENGINEERING PHYSICS I; Subject: Physics; University: University of Texas - Austin; Term: Spring 2011;

Typology: Quizzes

2010/2011

Uploaded on 10/18/2011

nikkanhat
nikkanhat 🇺🇸

1 document

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Version 036/AACBA QTest #3 Antoniewicz (57420) 1
This print-out should have 4 questions.
Multiple-choice questions may continue on
the next column or page find all choices
before answering.
001 (part 1 of 4) 10.0 points
A plane is flying horizontally with speed
287 m/s at a height 1870 m above the ground,
when a package is dropped from the plane.
The acceleration of gravity is 9.8 m/s2.
Neglecting air resistance, when the package
hits the ground, the plane will be
1. behind the package.
2. directly above the package. correct
3. ahead of the package.
Explanation:
Both have the same velocity at the time of
release. Gravitational acceleration does not
change horizontal velocity, so the plane will
be directly above the package.
002 (part 2 of 4) 10.0 points
What is the horizontal distance from the re-
lease point to the impact point?
1. 4596.55
2. 6641.95
3. 5606.67
4. 6237.66
5. 6300.31
6. 7180.97
7. 5067.8
8. 5667.51
9. 4924.2
10. 2621.08
Correct answer: 5606.67 m.
Explanation:
For the vertical fall, h=1
2g t2, or t=
s2h
g. The horizontal distance traveled is
x=v t
=vs2h
g
= (287 m/s)s2 (1870 m)
9.8 m/s2
= 5606.67 m .
003 (part 3 of 4) 10.0 points
A second package is thrown downward from
the plane with a vertical speed v1= 72 m/s.
What is the magnitude of the total velocity
of the package at the moment it is thrown as
seen by an observer on the ground?
1. 286.472
2. 199.612
3. 278.677
4. 251.754
5. 295.894
6. 190.184
7. 184.765
8. 202.608
9. 176.706
10. 162.25
Correct answer: 295.894 m/s.
Explanation:
The velocity is the vector sum of the vertical
and horizontal components of velocity as seen
from the ground. Hence the scalar speed is
s=qv2+v2
1
=q(287 m/s)2+ 72 m/s2
= 295.894 m/s.
004 (part 4 of 4) 10.0 points
What horizontal distance is traveled by this
package?
1. 5694.25
2. 3490.01
3. 2228.61
4. 2134.05
5. 7478.24
6. 3881.49
7. 4481.76
8. 3387.94
9. 3638.33
10. 5162.98
pf2

Partial preview of the text

Download Solved Quiz 3 for Engineering Physics I | PHY 303K and more Quizzes Physics in PDF only on Docsity!

Version 036/AACBA – QTest #3 – Antoniewicz – (57420) 1

This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.

001 (part 1 of 4) 10.0 points A plane is flying horizontally with speed 287 m/s at a height 1870 m above the ground, when a package is dropped from the plane. The acceleration of gravity is 9.8 m/s^2. Neglecting air resistance, when the package hits the ground, the plane will be

  1. behind the package.
  2. directly above the package. correct
  3. ahead of the package.

Explanation: Both have the same velocity at the time of release. Gravitational acceleration does not change horizontal velocity, so the plane will be directly above the package.

002 (part 2 of 4) 10.0 points What is the horizontal distance from the re- lease point to the impact point?

Correct answer: 5606.67 m.

Explanation:

For the vertical fall, h =

g t^2 , or t = √ 2 h g

. The horizontal distance traveled is

x = v t

= v

2 h g

= (287 m/s)

2 (1870 m) 9 .8 m/s^2 = 5606.67 m.

003 (part 3 of 4) 10.0 points A second package is thrown downward from the plane with a vertical speed v 1 = 72 m/s. What is the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground?

Correct answer: 295.894 m/s.

Explanation: The velocity is the vector sum of the vertical and horizontal components of velocity as seen from the ground. Hence the scalar speed is

s =

v^2 + v^21

=

(287 m/s)^2 + 72 m/s^2 = 295.894 m/s.

004 (part 4 of 4) 10.0 points What horizontal distance is traveled by this package?

Version 036/AACBA – QTest #3 – Antoniewicz – (57420) 2

Correct answer: 3881.49 m.

Explanation: The time of the vertical fall is now deter- mined by

h = v 1 t +

g t^2

0 =

g t^2 + v 1 t − h

t =

−v 1 +

v^21 + 4 ( 12 ) g h 2 12 g

−v 1 +

v^21 + 2 g h g = 13.5243 s.

The horizontal distance is

x = v t = (287 m/s) (13.5243 s) = 3881.49 m.