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matrenka [14]
4 years ago
5

Which of two factors influence the weight of an object due to gravitational pull?

Physics
2 answers:
Novosadov [1.4K]4 years ago
8 0
Mass and distance is correct
Tresset [83]4 years ago
7 0
The masses of the object and the planet it's on, and the distance between their centers.
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long, straight, vertical wire carries a current upward. Due east of this wire, in what direction does the magnetic field point?
Kamila [148]

Answer:

The correct answer to the question is;

North

Explanation:

Fleming's left hand rule   states that the direction of a magnetic field acting on a current carrying conductor is perpendicular to both the force acting on the current carrying conductor and the direction of the current.

Therefore, since the direction of the current is upward and a force is acting with a direction towards the east of the wire then the direction of the magnetic field is northwards.

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3 years ago
How to find net force
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3 years ago
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A bungee jumper of mass 75kg is attached to a bungee cord of length L=35m. She walks off a platform (with no initial speed), reac
attashe74 [19]

Answer:

1. 77.31 N/m

2. 26.2 m/s

3. increase

Explanation:

1. According to the law of energy conservation, when she jumps from the bridge to the point of maximum stretch, her potential energy would be converted to elastics energy. Her kinetic energy at both of those points are 0 as speed at those points are 0.

Let g = 9.8 m/s2. And the point where the bungee ropes are stretched to maximum be ground 0 for potential energy. We have the following energy conservation equation

E_P = E_E

mgh = kx^2/2

where m = 75 kg is the mass of the jumper, h = 72 m is the vertical height from the jumping point to the lowest point, k (N/m) is the spring constant and x = 72 - 35 = 37 m is the length that the cord is stretched

75*9.8*72 = 37^2k/2

k = (75*9.8*72*2)/37^2 = 77.31 N/m

2. At 35 m below the platform, the cord isn't stretched, so there isn't any elastics energy, only potential energy converted to kinetics energy. This time let's use the 35m point as ground 0 for potential energy

mv^2/2 = mgH

where H = 35m this time due to the height difference between the jumping point and the point 35m below the platform

v^2/2 = gH

v = \sqrt{2gH} = \sqrt{2*9.8*35} = 26.2 m/s

3. If she jumps from her platform with a velocity, then her starting kinetic energy is no longer 0. The energy conservation equation would then be

E_P + E_k = E_E

So the elastics energy would increase, which would lengthen the maximum displacement of the cord

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Hopping is performed on 2 feet<br><br><br> True<br> False
gizmo_the_mogwai [7]
True welcome₩₩ sorry
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