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Nimfa-mama [501]
3 years ago
5

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

, the tension in the cable was 7000 N. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N drag force.
(a) What was the tension in the cable when the craft was being lowered to the seafloor?
(b) What was the tension in the cable when the craft was being raised from the seafloor?
Physics
1 answer:
rusak2 [61]3 years ago
6 0

Answer:

a) 5200 N  b) 8800 N

Explanation:

a) tension in the cable when it was being lowered to the sea floor = weight of the object which acts downward ( equals the tension in the cable when the craft was stationary in opposite direction)  - the drag force which will act upward =  7000 - 1800 = 5200 N

b) tension in the cable when the craft was being raised since the tension will act upward and the drag force and the weight will act downward = 7000 + 1800 = 8800 N

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An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 1 N, F2 = 9 N, and F3 = 5
hammer [34]

Answer:

solved

Explanation:

a) F_net = (F2 - F3)i - F1 j

b) |Fnet| = sqrt( (F2 - F3)^2 + F1^2)

= sqrt( (9- 5)^2 + 1^2)

= 4.123 N

c) θ = tan^-1( (Fnet_y/Fnet_x)

= tan^-1( -1/(9-5) )

= -14.036°

7 0
3 years ago
Is your increase in gravitational potential energy the same in both cases? When Climbing a mountain on a zigzag path and on a st
baherus [9]

Answer:

The increase in gravitational potential energy is the same in both cases

Explanation:

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gravitational potential energy = mgh ( same in both cases )

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7 0
3 years ago
In a long, straight, vertical lightning stroke, electrons move downward and positive ions move upward and constitute a current o
uranmaximum [27]

The number of revolutions the electron completes in 60.0-μs of the strike is 134.

A magnetic field, a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

Electrons go downward and positive ions move upward in a long, straight, vertical lightning stroke, creating a current of magnitude I = 20.0 kA.

A free electron travels through the air at a speed of v = 300 m/s at a place r = 50.0 m east of the stroke's center.

Let the magnetic field be B, and F be the magnetic force.

Counterclockwise horizontal arcs of field lines are produced by the upward lightning current.

We have, B = 8 × 10⁻⁵ T and;

The mass of an electron is, m = 9.11 × 10⁻³¹ kg

The time interval is Δt = 60 μs = 60 × 10⁻⁶

The angular frequency is given as:

ω = qB /m = 2πN / Δt

Where the number of revolutions is N.

So,

N = qBΔt /2πm

N = (l.60 × l0⁻¹⁹)(8 × l0⁻⁵)(60 × 10⁻⁶) / 2π(9.11 × 10⁻³¹ kg)

N = 134 revolutions

Learn more about current here:

brainly.com/question/1100341

#SPJ4

7 0
2 years ago
A security guard walks at a steady pace traveling 170 m in one trip around the perimeter of a building.
WARRIOR [948]
Speed= distance/time
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6 0
3 years ago
A thin, metallic spherical shell of radius 0.227 m has a total charge of 6.03 × 10 − 6 C placed on it.
KATRIN_1 [288]

Answer:

Explanation:

Given

radius r=0.227 m

Charge on surface Q=6.03\times 10^{-6} C

Point Charge inside sphere q=1.15\times 10^{-6} C

Electric Field at r=0.735 m

Treating Surface charge as Point charge and applying Gauss law

E_{total}A=\frac{q_{enclosed}}{\epsilon _0}

where A=surface area up to distance r

E_{total}=\frac{Q+q}{4\pi r^2}

E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}

E_{total}=1.194\times 10^{5} N/C

3 0
4 years ago
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