Answer:
a) t=1s
y = 10.1m
v=5.2m/s
b) t=1.5s
y =11.475 m
v=0.3m/s
c) t=2s
y =10.4 m
v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
As gravity opposes movement the sign in the equations is negative.:
y = vi*t - ½ g*t2 Equation 1
v=vit-g*t Equation 2
y: The vertical distance the ball moves at time t
vi: Initial speed
g= acceleration due to gravity
v= Speed the ball moves at time t
Known information
We know the following data:
Vi=15 m / s
t=1s ,1.5s,2s
Development of problem
We replace t in the equations (1) and (2)
a) t=1s
=15-4.9=10.1m
v=15-9.8*1 =15-9.8 =5.2m/s
b) t=1.5s
=22.5-11.025=11.475 m
v=15-9.8*1.5 =15-14.7=0.3m/s
c) t=2s
= 30-19.6=10.4 m
v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Think its Positive
hope this helpes
D. 51 N. The minimum applied force that will cause the television slide is 51 N.
In order to solve this problem we have to use the force of static friction equation Fs = μs*n, where μs is the coefficient of static friction, and n is the normal force m*g.
With μs = 0.35, and n = 15kg*9.8m/s² = 147 N
Fs = (0.35)(147 N)
Fs = 51.45 N
Fs ≅ 51 N
Although the sample is not shown in this question, we can conclude that it would be reasonably easy for David to provide evidence of the color, consistency, temperature, and texture of the soil.
<h3 />
<h3>What are these properties an example of?</h3>
These are all examples of the physical properties of a sample. Since we cannot see the sample that David is using, it would be safest to assume that he would have no trouble providing evidence as to the physical properties of the soil, the:
- Color
- Consistency
- Temperature
- Texture
are all examples of this.
Therefore, we can confirm that David can provide evidence of the color, consistency, temperature, and texture of the soil.
To learn more about physical properties visit:
brainly.com/question/24632287?referrer=searchResults
Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
