Explanation:
The given data is as follows.
F = 
q = 
v = 385 m/s
= 0.876
Now, we will calculate the magnitude of magnetic field as follows.
B = 
= 
= 
T
= 10.65 T
Thus, we can conclude that magnitude of the magnetic field is 10.65 T.
Characteristics help us to classify seeds because different plants have different features.
<h3>How are characteristics used to identify and classify plants?</h3>
The divisions classify plants that are based on whether they reproduce by spores or seeds. Spore-bearing plants include ferns, club mosses, and horsetail while on the other hand, Seed-bearing plants are divided into gymnosperms and angiosperms. Different plants have different characteristics and features so on the basis of these characteristics we can easily classify seeds whether they belong from angiosperm and gymnosperm.
So we can conclude that characteristics help us to classify seeds because different plants have different features.
Learn more about seeds here: brainly.com/question/18799172
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Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
Answer:
2Ω
Explanation:
If a 18Ω resistance is cut into three equal parts each of the resistance will be 18Ω/3 = 6Ω
Equivalent ratio in parallel is expressed as:
1/R = 1/6 + 1/6 + 1/6
1/R = 3/6
Cross multiply
3R = 6
R = 6/3
R = 2Ω
Hence the required equivalent resistance is 2Ω
