Heated air molecules in a hot air balloon soon carried thermal energy to the top of the balloon. this is an example of...

a small table has a mass of 4kg, stands on four legs, each leg having an area of 0.001 m2. what is the pressure exerted by the t

Gemiola [76]

Answer:

P = 10 kPa

Explanation:

Given that,

The mass of a small table, m = 4 kg

The area of each leg = 0.001 m²

We need to find the pressure exerted by the table on the floor. Pressure is equal to the force per unit area. So

$P=\dfrac{mg}{4\times A}\\\\P=\dfrac{4\times 10}{4\times 0.001}\\P=10000\ Pa\\\\or\\\\P=10\ kPa$

So, the required pressure is 10 kPa.

3 0

3 years ago

Sound waves have relatively long wavelengths. We can hear people around the corner before we can see them. Which wave behavior d

pogonyaev

I'd say diffraction since sound waves can bend around objects like corners. Let's say you're in the hallway and you can hear sound coming from a door. The sound waves diffract around the door and spread out into the hallway, making it possible for you to hear.
Also, you can hear it before you see it because light waves are shorter than sound waves and hardly diffract around doors.

6 0

3 years ago

A symbolic model for learning is a model that is observed in person.

kap26 [50]

Answer:

True

Explanation:

3 0

3 years ago

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Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6

Nana76 [90]

Answer:

The answer to the question is as follows

The acceleration due to gravity for low for orbit is 9.231 m/s²

Explanation:

The gravitational force is given as

$F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }$

Where $F_{G}$ = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹ $\frac{Nm^{2} }{kg^{2} }$

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by $F_{G}$ and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and $F_{G}$ = $\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }$ = 9.231 N

Therefore the acceleration due to gravity = $F_{G}$ /mass

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²

4 0

3 years ago

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A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. then the operator turns on the rid

ioda

<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116 So coefficient of static friction = 0.6116</span>

4 0

3 years ago