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NISA [10]
3 years ago
9

A measurement that has both magnitude and direction

Physics
1 answer:
Korvikt [17]3 years ago
7 0
<span>A measurement that both magnitude and direction is a vector quantity. An example of this is a moving car. The car exerts force due to its thrust and weight that runs in it. This will give us the magnitude of the car. The resulting motion of the car in terms of displacement, velocity and acceleration that determines its direction makes it a vector quantity. On the other hand, a measurement that has only magnitude is a scalar quantity. The energy exerted by the engine of the car is a scalar quantity.</span>
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The angle of reflection is the angle the ____ to the reflecting surface.
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Answer:

The angle of reflection is the angle the reflected rays make with a perpendicular line to the reflecting surface.  

Explanation:

Reflection It is the change of direction suffered by a luminous ray when hitting the surface of an object. The angle of reflection is that which is formed by the reflected ray and the normal vector to the study surface

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05. The time required to complete one lap around a perfectly circular track having a radius of 1,835 meters is 86
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Answer:

v = 134.06 m/s

Explanation:

Given that,

Radius of a circular track is 1,835 m

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v=d/t

On circular path,

v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 1835}{86}\\\\v=134.06\ m/s

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3 years ago
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
KengaRu [80]

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

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resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m

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3 years ago
A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int
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Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of

<em>F</em> = <em>m a</em>   →   <em>a</em> = <em>F </em>/ <em>m</em>

so that the cart's final speed is

<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>

<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>

<em />

If we force is halved, so is the accleration:

<em>a</em> = <em>F</em> / <em>m</em>   →   <em>a</em>/2 = <em>F</em> / (2<em>m</em>)

So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give

(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>

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3 years ago
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