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Anna71 [15]
2 years ago
6

How does the tension in your arms compare when you let yourself dangle motionless by both arms and by one arm

Physics
1 answer:
PolarNik [594]2 years ago
6 0

When you support yourself with two arms, the tension in each arm is half of the tension you experience when you support your weight with only one arm.

The tension in your arm is directly proportional to the weight of your body.

T = W = mg

When you support your weight with your two arms;

  • the upward force balancing the downward force due to the weight of your body will be distributed equally in both arms.

Tension \ in \ each \ arm = \frac{Total \ weight \ of \ your \ body}{2}

When you support the weight of your body with one arm,

  • the upward force balancing the downward force due to your weight will be on only one arm

Tension \ in \ the \ one \ arm = Total \ weight \ of \ your \ body

Thus, when you support yourself with two arms, the tension in each arm is half of the tension you experience when you support your weight with only one arm.

Learn more here: brainly.com/question/13443419

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olchik [2.2K]
Conductors (something that allows electricity to flow easily) allow for electricity to flow easily. This would be the wires. If you don't have conductors, then you cannot have electricity flow.

Insulators (something that doesn't allow electricity to flow through it) is important because it allows us to be able to touch the cables or place them next to one another and not shock ourselves

Hope this helps
6 0
4 years ago
Read 2 more answers
Show explicitly that ▽ . B-0 near a long straight wire carrying a current I. HINT: you may use Cartesian coordinates, but anothe
aivan3 [116]

Answer:

\bigtriangledown.B=0 is proved.

Explanation:

The magnetic field in the long current carrying wire is,

B=\frac{\mu_{0}I }{2\pi r } \phi

Here, I is the current, B is the magnetic field.

Now, by using cylindrical coordinates for the divergence of B.

\bigtriangledown.B=\frac{1}{s} \frac{d}{d\phi} B

Put the value of B in above equation.

\bigtriangledown.B=\frac{1}{s} \frac{d}{d\phi}(\frac{\mu_{0}I }{2\pi r } \phi)\\\bigtriangledown.B=0

Hence, it is prove that for a long current I carrying wire magnetic field divergence that is \bigtriangledown.B=0.

7 0
3 years ago
Determine the initial velocity of the ball if it reaches a height of 15 meters.
jasenka [17]

Answer:

the initial velocity of the  ball is 17.14 m/s

Explanation:

Given;

maximum height reached by the ball, h = 15 m

let the initial velocity of the ball = u

at maximum height, the final velocity of the ball is zero, v = 0

The initial velocity of the ball is calculated by using the following upward motion kinematic equation;

v² = u² - 2gh

0 = u² - 2(9.8 x 15)

u² = 294

u = √294

u = 17.14 m/s

Therefore, the initial velocity of the  ball is 17.14 m/s

4 0
3 years ago
Write your answer to the question below.
ElenaW [278]

Answer:

Hi... Potential energy is converted to kinetic energy and kinetic energy is converted to potential energy

8 0
3 years ago
The current required to stimulate the heart during ventricular fibrillation is about 110mA (1000mA = 1A). Assuming that a conduc
algol [13]

Answer:

Explanation:

Given

current required I=110 mA

Internal Resistance R=300 \Omega

According to ohm law

Current flows in a conductor is directly Proportional to the voltage applied.

V\propto I

V=IR

V=110\times 10^{-3}\times 300

V=33 V  

5 0
3 years ago
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