<u>ALL of the following work assumes NO AIR RESISTANCE:</u>
1). an object moving under the influence of only gravity, and not in orbit; its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²
2). a parabola
3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.
4). a). the one that was thrown horizontally; b). both hit the ground at the same time; c). both hit the ground with the same vertical velocity
5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion
6). a). 1.8 seconds; b). 13.1 meters; c). 17.6 m/s down; d). 7.3 m/s; gravity has no effect on horizontal motion
7). 45 m/s
8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand
9). a). 4.49 m/s; b). 29.7 m/s
10). 7.24 meters
11). 700 meters
12). A). 103.7 meters ( ! she's in big trouble ! ); B). 17.5 meters
Answer:
a) E = -4 10² N / C
, b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰
Explanation:
For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball
X axis
-
= m a
Axis y
- W = 0
Initially the system is in equilibrium, so zero acceleration
Fe =
T_{y} = W
Let us search with trigonometry the components of the tendency
cos θ = T_{y} / T
sin θ =
/ T
T_{y} = cos θ
= T sin θ
We replace
q E = T sin θ
mg = T cosθ
a) the electric force is
= q E
E =
/ q
E = -0.032 / 80 10⁻⁶
E = -4 10² N / C
b) the distance to this point can be found by dividing the two equations
q E / mg = tan θ
θ = tan⁻¹ qE / mg
Let's calculate
θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)
θ = tan⁻¹ 0.3265
θ = 18
⁰
sin 18 = x/0.30
x =0.30 sin 18
x = 0.093 m
c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations
X axis
= m aₓ
aₓ = q E / m
aₓ = 80 10⁻⁶ 4 10² / 0.01
aₓ = 3.2 m / s²
Axis y
W = m
a_{y} = g
a_{y} = 9.8 m/s²
The total acceleration is can be found using Pythagoras' theorem
a = √ aₓ² + a_{y}²
a = √ 3.2² + 9.8²
a = 10.31 m / s²
The Angle meet him with trigonometry
tan θ = a_{y} / aₓ
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-9.8) / 3.2
θ = -71.9⁰
Movement is two-dimensional type with acceleration in both axes
Answer:
<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>
Explanation:
<u>Operations With Vectors</u>
Given two vectors in rectangular components:

The sum of the vectors is:

The difference between the vectors is:

The magnitude of
is:

The angle
makes with the horizontal positive direction is:

The question provides the vectors:



Calculate:

The magnitude of
is:

The angle is calculated by:

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.
The magnitude of vector d is 16 and the angle with the x-axis is 270°
Try This Method(substitute your values)
Centripetal force = Mass X Velocity^2/ Radius
<span>F= mv^2/r </span>
<span>so 80.5 X 40^2/200 </span>
<span>=644N</span>