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Tanya [424]
3 years ago
13

Consider a string of length 1.0 meter, fixed at both ends, with mass 100 grams and tension 100 newtons. part a give the number o

f nodes and antinodes set up on the string if it is driven at a frequency of 95 hertz. be sure to include the ends of the string in your count. give the number of nodes followed by the number of antinodes, separated by a comma.
Physics
1 answer:
Bond [772]3 years ago
4 0
To answer the problem we would be using this formula which isv = sqrt(T/(m/L)) 
v = sqrt(100 N / [(0.100 kg)/(1.0 m)]) 
v = 31.62 m/s 
v = fλ 
31.62 m/s = (95 Hz)(λ) 
λ = 0.333 m 
For every wavelength along a string there will be 2 antinodes. 
1.0 m / 0.333 m = 3 
3 * 2 = 6 antinodes 
6 + 1 = 7 nodes
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Match: Substance ....... Definition Lewis base A. Provides H in water Bronsted-Lowry base B. Provides OH - in water Arrhenius ac
algol13

Answer:

A. Provides H in water - Arrhenius acid

B. Provides OH - in water - Arrhenius base

C. Proton donor - Bronsted-Lowry acid

D. Proton acceptor - Bronsted-Lowry base

E. Electron pair donor - Lewis base

F. Electron pair acceptor - Lewis acid.

Explanation:

Definition:

A Lewis base is any substance, such as the OH- ion, that can donate a pair of nonbonding electrons. A Lewis base is therefore an electron-pair donor.

A Bronsted-Lowry acid is a chemical species that donates one or more hydrogen ions in a reaction. Also, a Bronsted-Lowry base accepts hydrogen ions.

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons. In other words, it increases the number of H+ ions in the water. Also, an Arrhenius base dissociates in water to form hydroxide ions, OH-.

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4 years ago
Radio waves are propagated through the interaction of:.
cluponka [151]

Answer:

Electric and magnetic fields:

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a changing magnetic field creates a changing electric field      and

a changing electric field creates a changing magnetic field

5 0
3 years ago
A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.
Andrei [34K]

Answer:

v_{y}=35.21m/s

Explanation:

From the exercise we know the cannonball's <u>initial velocity</u>, the <u>angle</u> which its released with respect to the horizontal and its <u>initial height</u>

v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m

If we want to know whats the <u>y-component of velocity</u> we need to use the following formula:

v_{y}^2=v_{oy}^2+ag(y-y_{o})

Knowing that g=-9.8m/s^2

v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s

So, the cannonball's y-component of velocity is v_{y}=35.21m/s

6 0
3 years ago
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atroni [7]

Answer:

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Explanation:

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3 years ago
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Ulleksa [173]

Answer:

Benjamin

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3 years ago
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