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lbvjy [14]
3 years ago
15

I mostly know how to do these vector problems i’m just confused on how many decimal places i use

Physics
2 answers:
Firlakuza [10]3 years ago
7 0

significant figures rules!! since all of the problems use 2 sig figs, you would make your answer 2 as well.

Degger [83]3 years ago
5 0
I believe you use 4? I’m really unsure about that but I like your Joy Division picture
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The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?
belka [17]

Answer:

B) 0.3Hz

Explanation:

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A. In one short sentence, explain why we call the force of gravity an attractive force.
kondor19780726 [428]

Answer:

Explanation:

(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.

(b) Yes, it does and the formula for force of gravity between any two object is

F = G\frac{m1m2}{r}

where m1 and m2 are masses of the first and second object respectively

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How much protein would be consumed with a 3-ounce serving of
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15kg

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Justify why does a bird flap its wing to fly higher??​
SpyIntel [72]

Answer:

they use thermals and air currents to glide.

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3 years ago
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0
3 years ago
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