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3 years ago

At what point does the ball have the most gravitational potential energy?

Physics

1 answer:

nevsk [136]3 years ago

3 0

Point C. at this point, which is the highest point, all of the ball’s energy is gravitational potential energy.

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Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%

djyliett [7]

Answer:

$6.8370869499\times 10^{20}\ N$

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

e = Charge of electron = $1.6\times 10^{-19}\ C$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

$q=imbalance\times n\times e$

Total number of protons and electrons in each sphere

$n=\dfrac{mN_AZe}{M}$

$q=imbalance\times \dfrac{mN_AZe}{M}$

$q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C$

$q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C$

Electrical force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N$

The electrostatic force of attraction between them is $6.8370869499\times 10^{20}\ N$

4 0

3 years ago

Reducing, reusing, and recycling in your office is likely to _______.

iris [78.8K]

Conserve natural resources, energy and landfill space.

4 0

4 years ago

An object of mass 6.00 kg falls with an aceleration of 8.00 m/s2. The magnitude of air resitance must be ____ N

allsm [11]

Using Newton's Second Law, we can find the air resistance. We know the net force is equal to mass times acceleration. $F_{net} = m*a = (6.00kg)(8.00 \frac{m}{s^2}) = 48N 

F_{g} - F_{d} = 48N

48N = (6.00kg)(9.81m/s^2) - F_{d} 

F_{d} = 10.86N$

7 0

3 years ago

Two waves overlap producing a new wave that exhibits destructive

suter [353]

Answer:B

Explanation:

8 0

3 years ago

Read 2 more answers

A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, $\lambda = 1\ nC = 1\times 10^{- 9}\ C$

Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

$\lambda$ = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

7 0

3 years ago