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Stels [109]
3 years ago
10

A gas always spreads out to fill all available space. True or False Science Hw

Physics
2 answers:
Ira Lisetskai [31]3 years ago
7 0

This is True. Gases spread around freely and take up any available space. For example, there is oxygen. Oxygen is a gas we breathe in and spreads around freely, taking up any space.

GREYUIT [131]3 years ago
6 0

Answer:

true

Explanation:

i know this im in 6th grade

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A 150-newton force, applied to a wooden crate at an angle of 30° above the horizontal, causesthe crate to travel at constant vel
omeli [17]

Answer:

Explanation:

A component of 150 N in vertical direction will reduce the magnitude of reaction force.

reaction force exerted by the floor

= mg - 150 sin 30

where m is mass of the crate .

the magnitude of the horizontal component of the 150-newton force

150 cos30

= 130 N

This force tries to pull the crate in forward direction with acceleration but it has no acceleration . It is so because frictional force of equal magnitude acts on it in opposite direction which makes the net force acting on it equal to zero.

Hence frictional force is equal to 150 cos 30.

= 130 N .

3 0
3 years ago
9. A plane starts at rest & accelerates along the ground before takeoff. It
Phoenix [80]

Answer:

  9.877 m/s^2

Explanation:

The acceleration can be computed from ...

  d = (1/2)at^2

  (1600 m) = (1/2)a(18 s)^2

  a = (1600/162) m/s^2 ≈ 9.877 m/s^2

6 0
3 years ago
A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What
iogann1982 [59]

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

a=\omega^2 r

At the instant t = 0,

\omega=3.93 rad/s

So the centripetal acceleration of the tip of the blades is

a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2

5 0
3 years ago
PLEASE HELP. NOTICE HOW THEY PUT FUTURE
MrMuchimi

I think its <u>B</u> because it looks like it might indicate a future rain storm.

6 0
3 years ago
Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t
Musya8 [376]

Answer:

n = 1.266\times 10^{12}

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is \theta = 13 degree

thickness of string  300 mm = 0.3 m

sin\theta =\frac{2}{0.3 m}

r =0.3 sin 13 = 0.067 m

Fe = \frac{ kq_1 q-2}{d^2}

Fe = \frac{kq^2}{(2r)^2} = mg tan\theta

q^2 =  mg tan\theta \frac{(2r)^2}{k}

    = 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}

q^2 = 4.10\times 10^{-14}

q = 2.026 \times 10^{-7} C

q = ne

n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}

n = 1.266\times 10^{12}

3 0
3 years ago
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