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2 years ago

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An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the

new acceleration would be _____ m/s/s.

1 answer:

alexandr402 [8]2 years ago

5 0

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

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A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n

Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is its frequency

For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is

$\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m$

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

$f=\frac{v}{\lambda}$

where for the dolphin:

v = 1531 m/s is the wave speed

$\lambda=1.50 cm=0.015 m$ is the wavelength

Substituting into the equation,

$f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz$

C. 13.6 m

Again, the wavelength is given by:

$\lambda=\frac{v}{f}$

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

$\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m$

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

$\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m$

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

$\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m$

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

$\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m$

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

$f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz$

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2 years ago

Which amplitude of the following longitudinal waves has the greatest energy?

Rashid [163]

Which amplitude of the following longitudinal waves has the greatest energy?

amplitude = 10 cm; wavelength = 6 cm; period = 4 seconds

8 0

2 years ago

Read 2 more answers

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

2 years ago

A sphere of radius 5.00 cm carries charge 3.00 nC. Calculate the electric-field magnitude at a distance 4.00 cm from the center

OlgaM077 [116]

Answer:

a) E = 8.63 10³ N /C, E = 7.49 10³ N/C

b) E= 0 N/C, E = 7.49 10³ N/C

Explanation:

a) For this exercise we can use Gauss's law

Ф = ∫ E. dA = $q_{int}$ /ε₀

We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product

The area of a sphere is

A = 4π r²

if we use the concept of density

ρ = q_{int} / V

q_{int} = ρ V

the volume of the sphere is

V = 4/3 π r³

we substitute

E 4π r² = ρ (4/3 π r³) /ε₀

E = ρ r / 3ε₀

the density is

ρ = Q / V

V = 4/3 π a³

E = Q 3 / (4π a³) r / 3ε₀

k = 1 / 4π ε₀

E = k Q r / a³

let's calculate

for r = 4.00cm = 0.04m

E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³

E = 8.63 10³ N / c

for r = 6.00 cm

in this case the gaussine surface is outside the sphere, so all the charge is inside

E (4π r²) = Q /ε₀

E = k q / r²

let's calculate

E = 8.99 10⁹ 3 10⁻⁹ / 0.06²

E = 7.49 10³ N/C

b) We repeat in calculation for a conducting sphere.

For r = 4 cm

In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.

E = 0

In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is

E = k q / r²

E = 7.49 10³ N / C

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2 years ago

Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a

mote1985 [20]

Answer:

E=0

Explanation:

Electric field due to each thin sheet of charge=\sigma/2\varepsilon

let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .

In the region between the plates,the electric field due to each plate is in same direction,

E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=\sigma/\varepsilon

in the region outside the plates, the field due to the plates is in opposite directions

E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=-\sigma/2\varepsilon+\sigma/2\varepsilon

E=0

4 0

1 year ago