For the object that has a horizontal velocity component of 6 m/s and a vertical velocity component of 4 m/s we have:
1. The velocity of the object is 7.21 m/s.
2. The angle it makes with the horizontal is 33.7°.
1. The velocity of the object can be found as follows:
Where:
: is the horizontal component of the velocity = 6 m/s
: is the vertical component of the velocity = 4 m/s
Hence, the <u>velocity is</u>:
2. The angle it makes with the <u>horizontal </u>can be calculated with the following trigonometric function:
Where:
θ: is the angle it makes with the horizontal
Therefore, the <u>angle is</u>:
You can learn more about the components of the velocity here: brainly.com/question/2285233?referrer=searchResults
I hope it helps you!
Answer:
I = 0.002593 A = 2.593 mA
Explanation:
Current density = J = (3.00 × 10⁸)r² = Br²
B = (3.00 × 10⁸) (for ease of calculations)
The current through outer section is given by
I = ∫ J dA
The elemental Area for the wire,
dA = 2πr dr
I = ∫ Br² (2πr dr)
I = ∫ 2Bπ r³ dr
I = 2Bπ ∫ r³ dr
I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]
I = (Bπ/2) [R⁴ - (0.9R)⁴]
I = (Bπ/2) [R⁴ - 0.6561R⁴]
I = (Bπ/2) (0.3439R⁴)
I = (Bπ) (0.17195R⁴)
Recall B = (3.00 × 10⁸)
R = 2.00 mm = 0.002 m
I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]
I = 0.0025929449 A = 0.002593 A = 2.593 mA
Hope this Helps!!!
The density of the metal can be determined through the formula [n*MW]/ Na*[a^3] . substituting, we get,
<span>d = [n*MW]/ Na*[a^3]
</span><span>d = [4 atoms*42.3 g/mol]/ [6.022 x 1023atoms/mol* (sqrt 8 *1.20x10-10)^3]
</span>d = 0.719 g/cm3
Answer: Explanation:
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