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iVinArrow [24]
2 years ago
13

A vector has a component of 10 m in the x-direction, a component of 10 m in the y-direction, and a component of 5 m in the z dir

ection. The magnitude of this vector is?
Physics
1 answer:
Kobotan [32]2 years ago
8 0

The magnitude of this vector is 15

A vector is a quantity or phenomenon that has two independent properties: magnitude and direction. The term also denotes the mathematical or geometrical representation of such a quantity. Examples of vectors in nature are velocity, momentum, force, electromagnetic fields, and weight.

The magnitude of a vector formula is used to calculate the length for a given vector (say v) and is denoted as |v|. So basically, this quantity is the length between the initial point and endpoint of the vector.

Let vector be = a

component of vector in x direction = 10 i

component of vector in y direction = 10 j

component of vector in z direction = 5 z

vector a = 10 i + 10 j + 5 z

magnitude of vector a = |a| = \sqrt{10^{2} +10^{2} + 5^{2}    }

                                             = 15

To learn more about vector here

brainly.com/question/24256726

#SPJ4

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If a ball of radius 0.1 m is suspended in water, density = 997 kg/m^3, what is the volume of water displaced and the buoyant for
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Answer:

Part A

The volume of water displaced is 4.1887902 × 10⁻³ m³

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Explanation:

From the question, we have;

The radius of the ball suspended (barely floating) in the water, r = 0.1 m

The density of the water, ρ = 997 kg/m³

Part A

The volume of the ball = The volume of a sphere = (4/3)·π·r³

∴ The volume of the ball = (4/3) × π × 0.1³ = 0.0041887902 m³ = 4.1887902 × 10⁻³ m³

Therefore;

The volume of water displaced, V = The volume of the ball = 4.1887902 × 10⁻³ m³

The volume of water displaced, V = 4.1887902 × 10⁻³ m³

Part B

The buoyant force = The weight of the water displaced = Mass of the water, m × The acceleration due to gravity, g

The buoyant force = m × g

Where;

g ≈ 9.8 m/s²

The mass of the water, m = ρ × V

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7 0
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A guitar string is supposed to have a fundamental frequency 256 Hz. It currently has a fundamental frequency 248 Hz when the str
umka2103 [35]

Answer:

The tension to bring the guitar string into tune is 372.95 Hz.

Explanation:

Given;

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The tension on the string to bring the guitar string into tune is calculated as;

v = \sqrt{\frac{T}{\mu} } \\\\f\lambda =  \sqrt{\frac{T}{\mu} } \\\\f^2\lambda^2 = \frac{T}{\mu} \\\\f^2 =  \frac{T}{\mu \lambda^2}\\\\let \ {\mu \lambda^2} = k\\\\f^2 =\frac{T}{k} \\\\k = \frac{T}{f^2} \\\\\frac{T_1}{f_1^2} = \frac{T_2}{f_2^2}\\\\T_2 = \frac{T_1 f_2^2}{f_1^2} \\\\T_2 =  \frac{350 \times  256^2}{248^2} \\\\T_2 = 372.95 \ Hz

Therefore, the tension to bring the guitar string into tune is 372.95 Hz.

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