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2 years ago

14

Sound waves have the ability to cause objects to vibrate. If a paperback book is placed near a speaker and the volume of the spe

aker is amplified, the book can be torn apart into small pieces.

1 answer:

erik [133]2 years ago

4 0

Answer:

yes with a lot of time it eventualy could be, but in short term no

Explanation:

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Which embedded molecule in the cell membrane moves large molecules into and out of the cell?

Nezavi [6.7K]

The correct answer is A.

The cell membrane consists of a phospholipid bilayer with embedded proteins. Sometimes molecules are just too big to easily flow across the plasma membrane or dissolve in the water so that they can be filtered through the cell membrane. In these cases , the cells must put out a little energy to help get molecules in and out of the cell.

The proteins embedded in the plasma membrane form channels through which other molecules can pass. Some proteins act as carriers, that is they are 'paid" in energy to let a molecule attach to itself and then transport that molecule inside the cell. This is called active transport.

5 0

3 years ago

Compared to a plane gravitational pull on the ground to 7 miles in sky is what

Citrus2011 [14]

When you're in an airplane that's 7 miles up off the ground, the strength of gravity plunges to only 99.6 percent of its strength all the way down on the ground. A big heavy person, who weighs 200 pounds down at the airport, weighs only 199 pounds 4.7 ounces in a plane at the altitude of 7 miles.

4 0

2 years ago

A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?

Olenka [21]

car starts from rest

$v_i = 0$

final speed attained by the car is

$v_f = 5 m/s$

acceleration of the car will be

$a = 3.6 m/s^2$

now the time to reach this final speed will be

$t = \frac{v_f - v_i}{a}$

$t = \frac{5 - 0}{3.6}$

$t = 1.39 s$

so it required 1.39 s to reach this final speed

6 0

3 years ago

A spherical weather balloon is filled with hydrogen until its radius is 4.40 m. Its total mass including the instruments it carr

ipn [44]

Answer:

4515.49484 N

4329.10484 N

Explanation:

r = Radius of balloon = 4.4 m

m = Mass of balloon with instruments = 19 kg

g = Acceleration due to gravity = 9.81 m/s²

Volume of balloon

$v=\frac{4}{3}\pi r^3\\\Rightarrow v=\frac{4}{3}\pi 4.4^3\\\Rightarrow v=356.8179\ m^3$

The Buoyant force = Weight of the air displaced

$F=\rho vg\\\Rightarrow F=1.29\times 356.8179\times 9.81\\\Rightarrow F=4515.49484\ N$

The buoyant force acting on the balloon is 4515.49484 N

Net force on the balloon

$F_n=F-W\\\Rightarrow F_n=4515.49484-19\times 9.81\\\Rightarrow F_n=4329.10484\ N$

The net force on the balloon is given by 4329.10484 N

As the balloon goes up the pressure outside reduces as the density of air decreases while the air pressure inside the balloon is high hence, the radius of the balloon tend to increase as it rises to higher altitude.

5 0

2 years ago

A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its

dlinn [17]

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

   <em>70 km per second per megaparsec</em>.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

   (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

      (150 million) x (1 / 3,262,000) x (70 km/sec) =

   <em>3,219 km/sec </em>in the direction away from us (rounded)

4 0

2 years ago