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Elis
1 month ago
9

Benjamin Franklin has convinced his hapless assistant Mike Piepan to participate in an experiment on electiricty. Ben has set up

lightning rod, the rod of which extends into his laboratory. MIke is suspended from the ceiling by an insulating rope. To one side of Mike is the end of end of the lightning rod, on the other side is a metal rod that is "grounded" (i.e. conducts electricity into the Earth, especially a "charge dump"). A bolt of lightning strikes the rod, giving it an enormous quantity of excess negative charge. Assuming tht no charge leaps through the air in the lab, explain in as much details as possible, what you think will happen to poor Mr. Piepan?
Physics
1 answer:
Softa [3K]1 month ago
5 0
Mike will get an electric shock. Explanation: The human body acts as a conductor of electricity. When lightning strikes the rod, it acquires a negative charge and immediately discharges this charge when contacting the ground through conductive materials. As a result, Mike is likely to endure a significant electric shock as the negative charge travels through his body to the other rod and into the ground, potentially leading to numbness or even loss of consciousness.
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Gregor Mendel finished his pea plant experiments in 1863 and published his results in 1866. However, few scientists saw Mendel's
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Individuals inherit one factor per trait from each parent, and a dominant factor can conceal the expression of a recessive one when both are present.
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1 month ago
When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the
Maru [3345]

Answer:

a

The value at a point inside is Zero

b

The electric field is E = 2.7*10^{6} \ N/C

Explanation:

We know from the problem that

The charge magnitude is q = 3.0 \mu C = 3.0 *10^{-6} \ C

The radius of the spherical ball is r = 5.0 \ mm = 0.005 \ m

According to Gauss’s law, the enclosed charge within a conductor is zero which indicates that the electric field within the spherical ball is zero

On the outside, the electric field around the spherical ball is mathematically expressed as

E = \frac{kq}{ a^2}

Here a denotes a point outside the spherical ball with its value of a = 10 \ cm = \frac{10}{100} = 0.1 \ m

and k represents Coulomb's constant, valued at

k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

=> E = \frac{ 3 *10^{-6} * 9*10^9 }{ (0.1)^2}

=> E = 2.7*10^{6} \ N/C

5 0
1 month ago
A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav
kicyunya [3294]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

  The locomotive's acceleration is 1.6 m/s^2

  The duration taken to pass the crossing is 2.4 seconds.

  We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

  When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 m/s^2.

   32 = 0 + 1.6 * t

    t = 20 seconds.

  Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

6 0
2 months ago
Read 2 more answers
In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as w_{h}.

We aim to determine the heart's weight for a human

Using the provided information

w_{h}=0.5\times w

Where, h = heart weight

w = human weight

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

The final weight of a human heart is 0.93 lbs.

8 0
2 months ago
A 55 kg gymnast wedges himself between two closely spaced vertical walls by pressing his hands and feet against the walls. Part
ValentinkaMS [3465]

answer:

Let frictional forces due to both hands and feets be "Ff" each(since its given that they all are equal), acting in upward direction( in opposite direction of supposed motion).\\
Then since there is no motion of gymnast thus net frictional force due to both hands and feets will be exactly balanced by the weight of the gymnast,\\ i.e\\
4f_{f}=weight =mg\\
f_{f}=\frac{55x9.8}{4}\\
=134.75N

5 0
2 months ago
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