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SSSSS [86.1K]
4 years ago
13

Two students grab a slinky and start waving it up and down. A third student counts the number of waves that pass by every second

and measures the distance between the wave peaks. This data is recorded and a graph is made to show the results.
In looking at the graph, what wave property is remaining constant over the experiment?

Physics
2 answers:
deff fn [24]4 years ago
4 0

Velocity = frequency * wavelength

v = fλ, Just pick any points on the graph for frequency f and corresponding λ. Taking the first red point at the top. λ = 6m, f = 1 Hz, v = 6 * 1, v = 6 m/s  


V = 6 M/S

LuckyWell [14K]4 years ago
4 0

Answer:

Velocity

Explanation:

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To calculate work done on an object, _____. A. multiply the force in the direction of motion by the distance the object moved B.
o-na [289]
I believe your answer would be B, hope it helps

6 0
3 years ago
Read 2 more answers
Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p
boyakko [2]

Answer:

The number of free electrons per cubic meter is 7.61\times 10^{28}\ m^{-3}

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, \sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}

Density of metal, \rho=10.5\ g/cm^3

Atomic weight, A = 107.87 g/mol

Let n is the number of  free electrons per cubic meter such that,

n=1.3\ N

n=1.3(\dfrac{\rho N_A}{A})

Where

\rho is the density of silver atom

N_A is the Avogadro number

A is the atomic weight of silver

n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})

n=7.61\times 10^{22}\ cm^{-3}

or

n=7.61\times 10^{28}\ m^{-3}

Hence, this is the required solution.

6 0
3 years ago
A 10-g bullet moving horizontally with a speed of 1.8 km/s strikes and passes through a 5.0-kg block initially at rest on a hori
UNO [17]

Answer:

6.4 J

Explanation:

m = mass of the bullet = 10 g = 0.010 kg

v = initial velocity of bullet before collision = 1.8 km/s = 1800 m/s

v'  = final velocity of the bullet after collision = 1 km/s = 1000 m/s

M = mass of the block = 5 kg

V = initial velocity of block before collision = 0 m/s

V'  = final velocity of the block after collision = ?

Using conservation of momentum

mv + MV = mv' + MV'

(0.010) (1800) + (5) (0) = (0.010) (1000) + (5) V'

V' = 1.6 m/s

Kinetic energy of the block after the collision is given as

KE = (0.5) M V'²

KE = (0.5) (5) (1.6)²

KE = 6.4 J

4 0
4 years ago
This time particle A starts from rest and accelerates to the right at 65.5 cm/s
FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

d = v_i t + \frac{1}{2}at^2

d = 0 + \frac{1}{2}(65.5)t^2

d_1 = 32.75 t^2 cm

Now we know that B moves with constant speed so in the same time B will move to another distance

d_2 = 44 \times t

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

d_1 = 349 + d_2

32.75 t^2 = 349 + 44 t

on solving above kinematics equation we have

t = 4 s

4 0
3 years ago
Consider two isolated, charged conducting spheres: a large sphere and a second smaller sphere with a radius 6 times smaller than
Troyanec [42]

Let the bigger sphere be sphere 1 and the let the smaller sphere be sphere 2. Rest of the answer is in the picture.

4 0
3 years ago
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