Lab7_Projectile_Motion(1)
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PHY 101 Lab 7: Projectile Motion
Activity 1
Data Table 1
Tria
l
Sphere
θ
a = 0.71 (9.8)sin θ
v
x =
√
2
a∆ x
t =
√
2
h
g
Calculate
d
Distance
x = v
x
t
(meters)
Actual
distance
(meters)
Percent
error
1
Metal
30°
-6.87 m/s²
-17.60 m/s
2.56 s
-45.06 m
-44.80 m
0.58%
2
Acrylic
30°
-6.87 m/s²
-17.60 m/s
2.56 s -45.06 m
-44.80 m
0.58%
3
Metal
+5°
-6.67 m/s²
-9.87 m/s
1.48 s
-14.61 m
-14.40 m
1.46%
4
Acrylic
+5°
-6.67 m/s²
-9.87 m/s
1.48 s
-14.61 m
-14.40 m
1.46%
5
Metal
+10°
-3.79 m/s²
-3.71 m/s
0.98 s
-3.65 m
-3.60 m
1.39%
6
Acrylic
+10°
-3.79 m/s²
-3.71 m/s
0.98 s
-3.65 m
-3.60 m
1.39%
1.
Did the sphere in the experiment always land exactly where predicted? If not, why was there a difference between the distance calculated and the distance measured?
In most cases, the sphere did not land exactly where predicted. There is often a difference between the distance calculated and the distance measured due to various factors like air resistance, imperfections in the
rolling surface, and slight variations in the angle and acceleration. Additionally, the theoretical calculations assume idealized conditions that may not perfectly match real-world situations.
2.
Why is it important to use the grooved ruler to ensure that the sphere leaves the table in a horizontal direction?
Using the grooved ruler is important because it helps ensure that the sphere leaves the table with a horizontal velocity. Without the grooved ruler, there might be lateral (sideways) motion or uneven rolling, which could affect the horizontal direction. The grooved ruler helps guide the sphere, ensuring that it starts with a consistent horizontal velocity, which is essential for accurate predictions and comparisons.
3.
If the same experiment were performed on the moon, what would be different? Several differences would occur if the experiment were conducted on the Moon:
The gravitational acceleration on the Moon is about 1/6th of that on Earth (approximately 1.62 m/s²). This lower acceleration would affect
the time of flight, acceleration, and horizontal velocity of the sphere.
With the lower gravitational acceleration, the sphere would stay in the air longer and travel farther horizontally before hitting the lunar surface.
Air resistance is negligible on the Moon, so that factor would not affect the experiment.
The angle of the incline and the materials used for the sphere and surfaces would still matter, but their effects would be adjusted according to the lunar conditions.
4.
What is different about the vertical component of the sphere’s velocity and the horizontal component of the sphere’s velocity once the sphere leaves the table?
Once the sphere leaves the table, the vertical component of its velocity is influenced by the force of gravity, and it accelerates downward at a constant rate (assuming no air resistance). This vertical velocity component increases as it falls. In contrast, the horizontal component of its velocity remains relatively constant because there are no horizontal forces acting on it (assuming no air resistance or horizontal forces). Thus, the horizontal component of velocity remains nearly the same throughout the horizontal motion.
5.
If the same experiment were repeated with the same angles, but from a taller table, how would the results change?
If the experiment were conducted from a taller table (assuming all other factors remain constant), several changes would occur:
The sphere would have more time to accelerate as it rolls down the incline, which would result in a higher horizontal velocity when it leaves the table.
The time of flight would be longer because the sphere would fall from a greater height, and it would hit the floor farther away from the table.
The actual distance traveled by the sphere on the floor would be greater than in the previous trials, assuming other factors like angle and acceleration remain unchanged.
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Related Questions
Solve for y, Ty, Tx, Vx, Vy'', Vx'' and V₁
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all must be in 3rd signi. digit pls
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What is the value of coefficients A and B?
Ix ₁
15-2
w
O 5/61,25/61
O 15/61,-35/61
O-5/61,-25/61
O 15/61, 35/61
72
VI
352
t
Vo
Ⓒ√x
Vo-A√x + BIX
41
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PLEASE ANSWER QUICKLY
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Direction: Solve the problem.
X2= 5000 m
X1 = Om
t = Os
ts= 5s
tz = 10s
DA = ?
ig = ?
XA = ?
A
vg =?
x = t) = t? -t + 2
aA =?
ag =?
B
v = = 3r* + ++t
1. Sub-atomic particles A and B are moving in a 100-meter dash line in 10s. Particle A moves such that its
displacement x in meters is given by the function x = (t? – t+ 2)meters. Meanwhile, particle B is moving at a
velocity v = (3r' + - t² + t) m/s. If both particles A and B started from rest at the same time, Compute the
following;
a. Average velocity v of particle A and particle B in m/s in 10s trip.
b. Displacement x of particle A and particle B from the starting point at t = 5s.
c. Velocity v of particle A and particle B at t = 5s
d. Acceleration a of particle A and particle B at t = 5s.
(Hint: wcelention is the de ri vative of velocity; ainst =
dv
dt
2. In a graphing paper, graph the function ix = (t² – t + 2) in displacement x vs time t graph (x vs t) showing
displacement x in the y-axis and time t in the x-axis. Then, graphically…
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?
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Quèstion 33
Choose the correct combination of the symbol and multiplication factor for prefix: Mega
M and 106
G and 109
T and 1012
P and 1015
Moving to another question will save this response.
TOSHIBA
F5
F6
F7
F8
RO
TO
F9
10
40
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Formulas:
Fnet=ma
Fg=mg
Ff=μmg
d=d0 + v0t + 1/2 at2
a=(v-v0)/t
v2=v02 + 2a(d-d0)
g=9.80m/s2
Thank you for helping me with my homework. It is due in one hour, so I hope somebody can help me soon. Also, it is a long question so I don't mind if you subtract multiple questions out of my account. Thank you so much.
2a. Joy is standing on a bathroom scale in an elevator. Yes, people are staring at her. Her mass is 50 kg. What does the scale read when the elevator is at rest? Include the unit.
2b. What does the scale read when the elevator accelerates upward from rest at 2m/s2?
2c. What does the scale read when the elevator moves up with a constant velocity of 10m/s?
3a. Many people are familiar with the fact that a rifle recoils when fired. A gunpowder explosion creates hot gases that expand outward allowing the rifle to push forward on the bullet. Compare the force on the rifle with the force on the bullet in terms of force direction and magnitude. This is a conceptual question. No…
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ill BITE_LV ?
12:50 PM
@マ 84%
A physics.area.lv
Question 4 of 7
What physical quantity is represented by
symbol h in equation
gt2
o Height of the device
O Disc outer radius
I = m: 2
- 1
2h
o Rope length
o Height of the
pendulum fall
Submit Answer
く
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3) The upward velocity of the rocket is measured with respect to time and the data is given in the
following table. Velocity vs time data for a rocket
Time, t(s)
Velocity, V (m/s)
105.7
8.
175.2
12
278.2
We wanted to approximate the velocity profile by
v(t=at*+bt+c,5sts12
Please construct the set of linear equation and solve the equation for the coefficients a, b and c in v (t).
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A0.8 (Homework)
For the graph of vx versus t shown, estimate the following:
i) What is the slope of the tangent line at t = 0.6 ?
ii) What is the slope of the tangent line at t = 1.0 ?
iii) What is the slope of the tangent line at t = 1.4 ?
2
-1-
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82. Find the scalar components of three-dimensional vectors Gand H in the following figure and write the vectors in vector
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60°
45°
G = 10.0
145
30°
H = 15.0
10
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Physics
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iven:
V = 20-(Horizontally)
Vyi = 0-,Required: the horizontal distance of the stone.
Vxi = 20.
1
Step1: y yi+Vyit-gt2
30%= 20 +0-
10 x t?
t= 2.0 seconds.
Step2: X = X, + vVxit = 0+ 20 x 2.0 = 40 meter.
- Correct Answer is A.
Problem (4):
A fire fighter 32.0 m away from a building, directs a stream of water from a fire hose at an angle
of 37 with speed of 40.0 m/s. At what height does the stream hit the building?
A.21.9 m
B. 22.5 m
C. 19.1 m
D. 20.9 m
Solution:
(X;%3D0
Xf = 32 m
%3D
Given:
Required: the height the stream will reach on the building.
Vi = 40-
0;=370
Vxi =V; Cos 0 = 40 x cos 37 = 31.9-
Step1:
Vyi = Vị sin 0 = 40 x sin 37 = 24.1-
%3D
%3D
%3D
%3D
Step2: X = X, + Vxit
%3D
32 =0+31.9 xt t%3D 1.0 second
Step3: y = y,+ Vyjt-gt
2812
3D0+
(24.1 x 1.0)-x 10 x 12)
3D19.1 meter.
3 Correct Answer is C..
59
S.
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