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Gre4nikov [31]
3 years ago
9

BRAINLEST TO SOMEONE WHO ACTUALLY ANSWERS AND FREE 100 POINTS TO REAL ANSWER 2. Using the average time, calculate the oscillatio

n period, or the time it took for one oscillation. Fill the averages into the orange cells in the data table. (2 points)
a. Reminder: the times in the data table are for five oscillations. The period of a pendulum is equal to one oscillation (one back and forth). So, divide the average time by five to get the time for
one oscillation.

Physics
2 answers:
andrezito [222]3 years ago
7 0

Answer:

IM SO SORRY :((

Explanation:

White raven [17]3 years ago
3 0
I’m so so so sorry! I will try!
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a= ∆v / t = vf - vi / t

Explanation: change in velocity or ∆v can be expressed as (vf - vi)

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3 years ago
____ charges are attract each other
timama [110]

Answer:

One positive and one negative

3 0
3 years ago
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What are two ways that machines can change the way that work is done?
algol [13]
Machines makes work easier by increasing the amount of force that is applied, and changing the direction in which the force is applied !! Hope it helped (p.s. I had this same question)
5 0
3 years ago
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a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s
Pepsi [2]

Answer:

\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a:  \\  \sf \implies 0 = 30 + ( - 1.5)(t) \\   \sf  \implies 0 = 30 - 1.5(t) \\  \sf \implies 30 - 1.5(t) = 0 \\  \\  \sf Subtract  \: 30  \: from  \: both  \: sides: \\  \sf \implies (30 -  \boxed{ \sf 30}) - 1.5(t) =  \boxed{ \sf  - 30} \\  \\  \sf 30 - 30 = 0 :  \\  \sf \implies  - 1.5(t) =  - 30 \\  \\  \sf Divide  \: both  \: sides \:  of \:  - 1.5(t) =  - 30 \: by \:  - 1.5 :  \\  \sf \implies  \frac{  - 1.5(t)}{ \boxed{ \sf - 1.5}}  =  \frac{ - 30}{ \boxed{ \sf -1.5 }}  \\  \\  \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}}  = 1 :  \\  \sf \implies t =  \frac{ - 30}{ - 1.5}  \\  \\   \sf  \frac{ - 30}{ - 1.5}  =  \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}}  = 20 :  \\  \sf  \implies t = 20 \: sec

So,

Time in which train will come to rest = 20 seconds

4 0
3 years ago
Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210
kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

Given

Wavelength of light (\lambda )=565nm \approx 565\times 10^{-9}m

Width of slit(b)=0.210

(a)Width of central maximum located 1.80m from slit

=\frac{2\lambda L}{b}

=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}

=9.685 mm

(b)Width of the first order bright fringe

Y_1=\frac{\lambda \times L}{b}

Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}

Y_1=4.84mm

5 0
3 years ago
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