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Svetlanka [38]
1 year ago
7

Consider a balloon of mass 0.030kg being inflated with a gas of density 0.54kg/m. What will be the volume of the balloon when it

just begins to rise in the air of density 1.29kg/m3. (g=10m/s2) Detailed Explanation Please
Physics
1 answer:
Nonamiya [84]1 year ago
7 0

the weight of the balloon is .030 * 10 = 0.3 N

the weight of the gas of volume v is 0.54*10 N

The lifting force of a volume of v m³ of displaced air is 1.29v N

so, we need

1.29*10*v = 0.3 + 0.54*10*v

or

1.29v = 0.03+0.54v

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Twist-on connectors without the spring-steel coils (plastic threads only) are suitable for making branch-circuit connections.
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if it is a plastic connector it wont work but if there is metal or steel it will work

Explanation:

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3 years ago
Bruh how did i get 3 warnings and 3 brainliests removed
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3 years ago
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A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is
Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
P = i^2R

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

P = (2.0^2)(50) = \boxed{\text{ B. }200 W}

7 0
2 years ago
The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is
astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

E = \frac{T}{P*sin(\theta)}=  \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C

E = 826 N/C (in three significant figures)

Therefore, the magnitude of this field is 826 N/C

6 0
3 years ago
A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p
a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h =  ¹/₂gt²

2h = gt²

t² = 2h/g

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s

4 0
3 years ago
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