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Lera25 [3.4K]
3 years ago
6

a hot air balloon weighing 30N is tied to the ground by a string to prevent from floating off the ground. The volume of the ball

oon is 20m³ and the density of air is 1.3kgm-³. find fhe force exerted by the rope on the balloon​

Physics
1 answer:
Lesechka [4]3 years ago
5 0

Answer:

The force exerted by the rope is FT = 225.06 [N]

Explanation:

In order to solve this problem we must use a static analysis, since Globe does not move. For a better understanding in solving this problem, a free body diagram with the forces acting on the globe is attached.

The buoyant force acts upward as it causes the balloon to tend to float, the weight of the balloon tends to lower the balloon and the downward tension force does not allow the balloon to float

The buoyant force is defined by the following equation:

FB = Ro*V*g

where:

FB = Buoyant force [N]

Ro = density of the air = 1.3 [kg/m^3]

V = volume of the balloon = 20 [m^3]

g = gravity acceleration = 9.81 [m/s^2]

FB = 1.3*20*9.81 = 255.06 [N]

Now we do a sum of forces equal to zero in the y-axis

FB - 30 - FT = 0

255.06 -30 = FT

FT = 225.06 [N]

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Which of the following changes would lead to an increase in the efficiency of
ladessa [460]

Answer:

b

Explanation:

doing same amount of work while increase in amount of heat released

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3 years ago
Is the magnitude of the acceleration of the center of mass of spool A greater than, less than, or equal to the magnitude of the
Anon25 [30]

Answer:

The acceleration of the centre of mass of spool A is equal to the magnitude of the acceleration of the centre of mass of spool B.

Explanation:

From the image attached, the description from the complete question shows that the two spools are of equal masses (same weight due to same acceleration due to gravity), have the same inextensible wire with negligible mass is attached to both of them over a frictionless pulley; meaning that the tension in the wire is the same on both ends.

And for the acceleration of both spools, we mention the net force.

The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.

For this system, the net force on either spool is exactly the same in magnitude because the net force is a difference between the only two forces acting on the spools; the tension in the wire and their similar respective weights.

With the net force and mass, for each spool equal, from

ΣF = ma, we get that a = ΣF/m

Meaning that the acceleration of the identical spools is equal also.

Hope this Helps!

6 0
3 years ago
An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and
Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is F_{ma}=BqV---(i)

The given formula to find the electric force is F_{el}=qE---(ii)

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

Bqv=qE\\\\v=\frac{E}{B}

v=\frac{187500}{0.125} \\\\v=15\times10^5m/s

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

F_{ce}=\frac{mv^2}{r}---(iii)

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

Therefore,  the particle's charge-to-mass ratio is 958.1\times10^5C/kg

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

8 0
4 years ago
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an
Dominik [7]

Answer:

position as a function of time is y = 0.05 × cos(9.9)t

Explanation:

given data

mass = 5 kg

length = 10 cm = 0.1 m

displaced = 5 cm

to find out

position as a function of time

solution

we will apply here equilibrium that is

mass × g = k × length

put here value and find k

k = \frac{5*9.8}{.01}

k = 490 N/m

and ω is

ω = \sqrt{\frac{k}{m} }

ω = \sqrt{\frac{490}{5} }

ω = 9.9

so here position w.r.t  time is

y = 0.05 × cosωt

y = 0.05 × cos(9.9)t

so position as a function of time is y = 0.05 × cos(9.9)t

8 0
3 years ago
Which of the following are not correctly balanced reactions?
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