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Tamiku [17]
3 years ago
9

A hover craft also known as a air cushion vehicle glides on a cushion of air allowing it to travel with equal is land or water.

The first commercial hovercraft to cross the English channel the V. A-3 had a average speed of 26.7 M/S. Its momentum at the speed was 4.8×10^4 KGM/S. What was the mass of the V. A-3?
Physics
1 answer:
dimaraw [331]3 years ago
4 0

Answer: M = 1797.75 kg

Explanation:

given parameters are;

speed V = 26.7 M/S.

momentum P = 4.8×10^4 KGM/S.

What was the mass of the V. A-3?

Momentum P is the product of mass and velocity. That is, P = MV

Substitute V and P into the formula

4.8×10^4 = 26.7 × M

Make M the subject of formula

M = 4.8×10^4/ 26.7

M = 1797.75 kg

Therefore, the mass of the V. A-3 was 1797.75 kg

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Blood cell : Eukaryotic cell
and
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Synthetic plastics are made by linking many simple carbon molecules together to form much larger molecules. This process is call
professor190 [17]

Answer:

A. polymerization

Explanation:

Synthetic plastics are made by linking many simple carbon molecules together to form much larger molecules. This process is called polymerization.

Synthetic or artifical giant molecules consists of synthetic polymers such as plastics, elastomers etc. They are made up of simple monomers which links to form the complex and giant structure.

Monomers are the simplest unit of polymers. Polymers have very great sizes. The size mkaes their structure quite complex. This makes the molecules more disposed in a regular pattern with respect to one another.

The complexity of structure and the attendant effects accounts for the properties and uses that makes synthetic molecules very unique. For example, plastics can be extruded as sheets, pipes and or moulded into other objects.

8 0
4 years ago
Two point charges 3q and −8q (with q > 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so
riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q  8q / L²

Q = 8q  (x² / L²)

so charge required = - 8q  (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

3 0
3 years ago
Beth moves a 15 N book 20 meters in 10 seconds. How much power was produced?
Mice21 [21]

Answer:

30 Watts

Explanation:

 Power = Work/Time

Work = Force*Distance

Power = Force * Distance / Time

Power = 15 N * 20 meters / 10 sec

Power = 30 Watts

5 0
3 years ago
The Hubble Space Telescope (HST) orbits 569,000m above Earth’s surface. Given that Earth’s mass is 5.97 × 10^24 kg and its radiu
Soloha48 [4]
Refer to the diagram shown below.

M = 5.97 x 10²⁴ kg, mass of the earth
h = 5.69 x 10⁵ m, height of HST above the earth's surface
R = 6.38 x 10⁶ m, radius of the earth

Note that
G = 6.67 x 10⁻¹¹ (N-m²)/kg², gravitational acceleration constant.
R + h = 6.38 x 10⁶ + 5.69 x 10⁵ = 6.949 x 10⁶ m

The force between the earth and HST is
F = (GMm)/(R+h)²

Let v = tangential velocity of the HST.

The centripetal force acting on HST is equal to F.
Therefore
m*[v²/(R+h)] = (GMm)(R+h)²
v² = (GM)/(R+h)
    = [(6.67 x 10⁻¹¹ (N-m²)/kg²)*(5.97 x 10²⁴ kg)]/(6.949 x 10⁶ m)
    = 5.7303 x 10⁷ (m/s)²
v = 7.5699 x 103 m/s

Answer
The tangential speed of HST is about 7,570 m/s 

3 0
4 years ago
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