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valentina_108 [34]
3 years ago
5

A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a

ir is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Aleks [24]3 years ago
3 0

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

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timurjin [86]
We calculate current from the formula:
I= \frac{q}{t} , where q is a electric charge transferred over time t 
Time should be converted to seconds:
1h 15 min= 75min= 4500s
I=\frac{90C}{4500s}=0,02A Result is in unit-Ampere
5 0
3 years ago
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ivolga24 [154]

Answer:

D.diamond

Explanation:

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5 0
3 years ago
If a single circular loop of wire carries a current of 62 A and produces a magnetic field at its center with a magnitude of 1.20
s2008m [1.1K]

Given Information:  

Current in loop = I = 62 A

Magnitude of magnetic field = B = 1.20x10⁻⁴ T

Required Information:  

Radius of the circular loop = r = ?  

Answer:  

Radius of the circular loop = 0.324 m

Explanation:  

In a circular loop of wire with radius r and carrying a current I  induces a magnetic field B which is given by

B = μ₀I/2r

Please note that for an infinitely straight long wire we use 2πr whereas for circular loop we use 2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space

Re-arranging the equation yields

r = μ₀I/2B

r = 4πx10⁻⁷*62/2*1.20x10⁻⁴

r = 0.324 m

Therefore, the radius of this circular loop is 0.324 m

3 0
3 years ago
An object is dropped from a​ tower, 400 ft above the ground. The​ object's height above ground t seconds after the fall is ​s(t)
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Answer: v= 160ft/s

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Explanation:

s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.

0=400-16t^2

400=16t^2

25=t^2

t=5s

ds/dt=v(t)=0-32t

dv/dt=a(t)=-32 constant(gravity)

v(t)=-32(5s)= -160ft/s negative sign is only showing direction

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Serggg [28]

i don't know sorry

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