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3 years ago

Please help with these 3 questions...thank you so much! They are about Physical Science and balanced reactions, thanks again!

Physics

1 answer:

stealth61 [152]3 years ago

6 0

Adian for one of the answers i think its 2Cr + 3Cl2 --> 2CrCl3 correct me if im wrong. None of these were on my quiz so im going to try to figure out the other two

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A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 103 ft3/min, meaning that about 1.85 ✕ 103 cubic feet

densk [106]

Answer:

0.8726 $m^3/s$

Explanation:

We are to convert 1.85 x $10^3$ $ft^3/min$ to $m^3/s$

First, let us convert the numerator from ft3 to m3

1 ft3 = 0.0283 m3

Hence,

1.85 x $10^3$ ft3 = 1.85 x $10^3$ x 0.0283 m3

= 52.355 m3

Now, let us convert the denominator from minutes to seconds

1 min = 60 sec

Therefore;

1.85 x $10^3$ $ft^3/min$ = 52.355/60 $m^3/s$

= 0.8726 $m^3/s$

7 0

3 years ago

Name three situations in which force is created. describe the cause of the force in each situation

Darina [25.2K]

One situation in which force is created is when an object is moving and a force is created to stop that movement. Second situation is when an object is moving circularly and a force is created to move it towards the middle of the circle. The third situation is when a force is created that goes in the same direction as an object that is in movement.

6 0

3 years ago

The ratio of translational energy and total energy of a rolling sphere is given by

Rufina [12.5K]

The answer to your question is 5:7

4 0

3 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$