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STALIN [3.7K]
3 years ago
14

Tara goes kayaking at sunrise. She rows 50 feet in 20 seconds. She transfers

Physics
1 answer:
Katarina [22]3 years ago
3 0
100J/20s= 5 joules per second

Or 5 watts
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If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the
Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0
3 years ago
Which trial’s cart has the greatest momentum at the bottom of the ramp?
Licemer1 [7]
The momentum of each cart is given by:
p=mv
where
m is the mass of the cart
v is its velocity (at the bottom of the ramp)

To answer the problem, let's calculate the momentum of each of the 4 carts:
1) p=(200 kg)(6.5 m/s)=1300 kg m/s
2) p=(220 kg)(5.0 m/s)=1100 kg m/s
3) p=(240 kg)(6.4 m/s)= 1536 kg m/s
4) p=(260 kg)(4.8 m/s)=1248 kg m/s

Therefore, the cart with greatest momentum is cart 3, so the right answer is
<span>- trial 3, because this trial has a large mass and a large velocity</span>
8 0
3 years ago
For the long life cells we have to connect them in ____ combination​
NISA [10]

Answer:

Parallel combination.

5 0
3 years ago
Read 2 more answers
Suppose we could shrink the earth without changing its mass..?At what fraction of its current radius would the free-fall acceler
Drupady [299]

Answer:

at R/\sqrt{3}

Explanation:

The free-fall acceleration at the surface of Earth is given by

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

The formula can be rewritten as

R=\sqrt{\frac{GM}{g}} (1)

We want to shrink the Earth at a radius R' such that the acceleration of gravity becomes 3 times the present value, so

g' = 3g

Keeping the mass constant, M, and substituting into the equation, we have

3g=\frac{GM}{R'^2}

R'=\sqrt{\frac{GM}{3g}}=\frac{1}{\sqrt{3}}\sqrt{\frac{GM}{g}}=\frac{R}{\sqrt{3}}

5 0
3 years ago
Read 2 more answers
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
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