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shusha [124]
3 years ago
7

If you pour different liquids into a graduated cylinder, which layer would be at the bottom?

Physics
1 answer:
klasskru [66]3 years ago
5 0

the one that is more dense!

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A typical jet airliner has a cruise airspeed of 900 km/h , which is its speed relative to the air through which it is flying. If
Luba_88 [7]

Explanation:

It is given that,

Speed of the jet airplane with respect to air, v_{PA} = 900\ km/h          

If the wind at the airliner’s cruise altitude is blowing at 100 km/h from west to east, v_{AG}= 100\ km/h

(A) Let v_{PG} is the speed of the airliner relative to the ground if the airplane is flying from west to east,

v_{PG}=900-100=800\ km/h

(B) Let v'_{PG} is the speed of the airliner relative to the ground if the airplane is flying from east to west,

v'_{PG}=900+100=1000\ km/h

Hence, this is the required solution.                                            

8 0
3 years ago
Which phase of cell division is shown?
dalvyx [7]

Answer:

what was the answer

Explanation:

im taking the quiz and could use some help

6 0
3 years ago
If a star appears to be very red (like Betelgeuse), it is likely to be in which spectral class?
Kipish [7]
It would be spectral class M

hope this helps


7 0
3 years ago
Read 2 more answers
A block of aluminum with a volume of 10 cm3 is placed in a beaker of water filled to the brim. Water overflows. The same is done
mihalych1998 [28]

Answer:

a) Both blocks displace the same amount of water.

Explanation:

Specific gravity (S.G.) is a quantity that tells how much and object is submerged on water and is given by:

S.G=\frac{\rho_{block}}{\rho_{fluid}}

with ρ the densities

Density of water is 997 \frac{kg}{m^{3}} and density of aluminum is 2712 \frac{kg}{m^{3}}

So:

S.G=\frac{2712}{997}=2.72

A S.G. value bigger than one means the object is totally submerged so the water displaced is equal to the volume of the cube.

For the lead (density = 11340\frac{kg}{m^{3}}) block we're going to calculate specific gravity in this case:

S.G=\frac{11340}{997}=11.37

Again S.G gravity is bigger than 1, so the lead block is totally submerged too, that implies the volume of water displaced is the volume of the block, that is the same volume as the aluminum block, so a is the correct one.

7 0
3 years ago
A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.796
Agata [3.3K]

Answer:

distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

Explanation:

Given that;

mass of block m = 0.200 kg

distance travelled d = 0.796 m

time t = 2.00 s

m₂ = 0.400 kg

If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

Now, using the second equation of motion;

d = ut + (\frac{1}{2} × at²)

as the object started from rest, u=0

so, we substitute

0.796  = 0×2 + (\frac{1}{2} × a(2)²)

0.796  = 0 + (\frac{1}{2} × 4a)

0.796  = 2a

a = 0.796 / 2

a = 0.398 m/s²

using first equation of motion

V_{f} = u + at

we substitute

V_{f} = 0 + 0.398 × 2

V_{f} = 0.796 m/s

now, average velocity is given as;

V_{avg} = ( 0.796 m/s  + 0 ) / 2

V_{avg} = ( 0.796 m/s  + 0 ) / 2

now, distance as the block moves in 2s will be;

D = [( 0.796 m/s  + 0 ) / 2 ] × 2

D = 0.796 m

Therefore, distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

5 0
3 years ago
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