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3 years ago

Muatan listrik yang mengalir dari potensial tinggi ke potensial rendah disebut

Physics

1 answer:

igomit [66]3 years ago

8 0

Answer:

Muatan listrik yang mengalir dari potensial tinggi ke potensial rendah disebut

elektron

Explanation:

semoga ini membantu (:

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On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed

JulsSmile [24]

Answer: 1.25 m

Explanation:

Given

initial velocity $(v_i )=3 m/s$

Final velocity $(v_f)=0.55\times 3=1.65 m/s$

Change in kinetic Energy =work done by Friction

change in Kinetic Energy $=\frac{m}{2}\left ( v_i^2-v_f^2\right )$

work done by friction $=\mu mgL$

$\frac{m}{2}\left ( 3^2-1.65^2\right )=0.25\cdot mg\times L$

$3.135=0.25\times 9.8\times L$

L=1.25

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3 years ago

Suppose that 4 charged particles, all having a positive charge Q, are placed and fixed at the vertices of a square, where every

algol [13]

Answer:

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3 years ago

The compound co2 is an example of a(n) atom element ionic compound covalent compound

alexandr402 [8]

Ionic compound is the answer

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3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

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3 years ago

Using Kepler's 3rd Law of Planetary motion, determine the distance between the Earth and the center of the Moon.

Ira Lisetskai [31]

Answer: the distance earth to moon is about 238,855 miles

Explanation:

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3 years ago