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3 years ago

9

A wave with a frequency of 190 hz and a wavelength of 28.0 cm is traveling along a cord. the maximum speed of particles on the c

ord is the same as the wave speed. part a what is the amplitude of the wave

1 answer:

raketka [301]3 years ago

8 0

Check the attached file for the answer.

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How might burning fossil fuels affect the composition of gases in the atmosphere?

elena-14-01-66 [18.8K]

No "might<span>". The amount of CO2 in the </span>atmosphere<span> HAS gone up since the start of industrialisation as the result of </span>burning fossil fuels<span>.</span>

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2 years ago

Why is it difficult to walk on muddy road?

Gnesinka [82]

Because of internal friction between layers of mud particles called viscosity. When you walk, your foot exerts a force on the mud; and according to Newton, the mud also (is supposed to) exert an equal opposite force, which leading to an equal net resultant force in your direction, propelling you forward.

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3 years ago

An object starts from rest and uniformly accelerates at a rate of 1.25 m/s2 for 7.0 seconds.

stich3 [128]

Explanation:

Since its accelerating, the velocity vs time graph is linear

For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula

A= (vf - vi)/t

a= vf-0/t

1.25=vf / 7

1.25*7=vf

8.75 = vf

Now for displacement plug all the values in

X = 1/2(vf-vi)/t formula

The displacement (x) is 30.625 m

For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above

The answer is t = 5.02

Im pretty sure all the answers are correct

8 0

2 years ago

What is the mass of a large ship that has a momentum of 1.60x10^9 kg*m/s and is moving at a velocity of 10m/s?

Elden [556K]

Answer:

160000000 kg.

Explanation:

p=mv

p=1.6x10^9

v=10m/s

rearrange and substitute:

(1.6x10^9)=m(10)

m=(1.6x10^9)/10

m= 1.6x10^8 kg.

7 0

2 years ago

the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t

givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

$a_{t} = v = 0.8 m/s^{2}$

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

= 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

= 0.8

at x= 5 Km

dy/dx = 0.8(5)

= 4

$p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }$

now insert the values,

$p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km$

→ Now the normal component of acceleration is given by

$a_{n} = \frac{v^{2} }{p}$

= (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

$a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}$

$a = [(0.8)^{2} + (0.457)^{2}]^{0.5}$

a = 0.921 m/s²

4 0

2 years ago