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Alja [10]
3 years ago
11

It is possible to hang from a bare power line and not get electrocuted as long as you are not touching the ground, or any conduc

tor that is touching the ground.
Physics
1 answer:
NNADVOKAT [17]3 years ago
3 0
Yes i believe so. depending on the covering of the line and other environmental confounding variables 
You might be interested in
A hula hoop is rolling along the ground with a translational speed of 26 ft/s. It rolls up a hill that is 16 ft high. Determine
nikklg [1K]

Answer:12.8 ft/s

Explanation:

Given

Speed of hoop v=26\ ft/s

height of top h=16\ ft

Initial energy at bottom is

E_b=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2

Where m=mass of hoop

I=moment of inertia of hoop

\omega=angular velocity

for pure rolling v=\omega R

I=mR^2

E_b=\frac{1}{2}mv^2+\frac{1}{2}mR^2\times (\frac{v}{R})^2

E_b=mv^2=m(26)^2=676m

Energy required to reach at top

E_T=mgh=m\times 32.2\times 16

E_T=512.2m

Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy

\Delta E=676m-512.2m=163.8m

Therefore

163.8 m=mv^2

v=\sqrt{163.8}

v=12.798\approx 12.8\ ft/s

7 0
3 years ago
Some of the energy the car gained as it was pulled was lost because of heat from friction. The rest of the energy was transforme
Lisa [10]

Answer:

454,320 joules

Explanation:

The work done on an object is equal to its change in kinetic energy: Change in KE = F × d.

Plug the values for F and d into the formula and solve:

Change in KE = 2,524 × 180

= 454,320 joules

The roller coaster gains 454,320 joules of energy from the work done on it by the chain.

7 0
4 years ago
Read 2 more answers
Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag
Salsk061 [2.6K]

Answer:

The resultant electric force is 14.8N to the right.

Explanation:

Since the three charges aren't in the same line, we have to break down the force in components. First, we need to know the distance from the third charge to the other ones. That is made using the Pythagorean Theorem. As the figure is symmetric with respect to the x-axis, the two distances are the same:

r=\sqrt{(0.50m)^{2}+(0.70m)^{2}}=0.86m

Now, we use the Coulomb's Law to obtain the magnitude of the individual forces caused by each charge on the third charge:

|F_{13}|=k\frac{q_1q_3}{r^{2}} \\\\|F_{13}|=(9*10^{9}Nm^{2}/C^{2})\frac{(2.5*10^{-9}C)(3.0*10^{-9}C)}{(0.86m)^{2}}\\\\|F_{13}|=9.1N

For the same reason the distances are the same, the magnitude of the forces are the same:

|F_{23}|=|F_{13}|=9.1N

So, to get the resultant force, we have to break down this forces in components. To do this, we need their angles with respect to the x-axis. Let θ₁ and θ₂ be these angles, respectively. Then, we calculate them using trigonometry:

\theta_1=\arctan(\frac{-0.50m}{0.70m})=-35.5\°\\\\\theta_2=\arctan(\frac{0.50m}{0.70m})=35.5\°

Now, we calculate the components of the forces:

F_{13}_x=F_{13}\cos\theta_1=9.1N\cos(-35.5\°)=7.4N\\\\F_{13}_y=F_{13}\sin\theta_1=9.1N\sin(-35.5\°)=-5.3N\\\\F_{23}_x=F_{23}\cos\theta_2=9.1N\cos(35.5\°)=7.4N\\\\F_{23}_y=F_{23}\sin\theta_2=9.1N\sin(35.5\°)=5.3N

Evidently, the y-components cancel out, and the resultant electric force on the third charge is 7.4N+7.4N=14.8N along the x-axis (to the right, because it's positive).

8 0
3 years ago
Please help I need this its for a grade
irina1246 [14]

Which sentence from paragraph 4 most precisely explain the factors that determine the force of gravity?

  • <em>"</em><em> </em><em>In</em><em> </em><em>other</em><em> </em><em>words</em><em>,</em><em> </em><em>these</em><em> </em><em>equations</em><em> </em><em>states</em><em> </em><em>the</em><em> </em><em>gravitational</em><em> </em><em>force</em><em> </em><em>between</em><em> </em><em>two</em><em> </em><em>bodies</em><em> </em><em>is</em><em> </em><em>proportional</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>product</em><em> </em><em>of</em><em> </em><em>their</em><em> </em><em>masses</em><em> </em><em>and</em><em> </em><em>inversely</em><em> </em><em>proportional</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>square</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>distance</em><em> </em><em>between</em><em> </em><em>them</em><em>.</em><em>"</em>

It's formula is g=GM/r²

4 0
3 years ago
What is the density of a rock that has a mass of 454g and volume of 100.0cm
Mademuasel [1]

Answer:

4.54 g/cm3

Explanation:

7 0
3 years ago
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