Answer:
Option B. O because the net force was 5 N in Alfredo's direction
Explanation:
To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:
Force of pull by Mason (Fₘ) = 15 N
Force of pull by Alfredo (Fₐ) = 20 N
Net force (Fₙ) =?
Fₙ = 20 – 15
Fₙ = 5 N in Alfredo's direction
From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.
Answer:
V = 10 km / 1 hr = 10 km/hr
V = -10 j km / hr if one were to use i, j, k as unit vectors with the usual orientation
Answer:
the answer is b
Explanation:
Second and third class levers are differentiated by <u>the location of the </u><u>load.</u>
<em>Hope</em><em> </em><em>this</em><em> </em><em>help</em><em> </em><em>you</em><em> </em><em>out </em><em>and have</em><em> </em><em>a </em><em>nice</em><em> </em><em>day </em><em>=</em><em>)</em>
Even though humans share 100% of the same genes, the instructions contained within the genes are not entirely identical. Each person is unique. People have different hair colors, facial structures, and other traits. These differences between individuals result from very small differences in their DNA sequences. DNA also contains many so-called "housekeeping genes" that control important metabolic processes. As you will see, some of the differences in these genes can cause illness.
Although the DNA of any two people on Earth is, in fact, 99.9% identical, even a tiny difference can have a big effect if this difference is located in a critical gene.
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so 
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
