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Artemon [7]
4 years ago
14

Two children of about the same weight are playing at the playground. They both climb up to the top of a small tower. One slides

back down to ground level on a frictionless slide. The other freely drops down from the tower. Both children start their trips back down to the ground with an initial speed of zero.
Which combination of statements below is most accurate? [Enter your answer as a string of the letter options you believe are correct. For instance, if you think options A, D, and E are correct, then enter "ADE"]

A) The work done by gravity on each child is positive as they climb the tower.
B) The children have different kinetic energies on reaching ground level
C) The work done by gravity on both children is negative as they descend from the tower.
D) The overall work done by gravity is zero.
E) None of the above
Physics
1 answer:
olya-2409 [2.1K]4 years ago
4 0

Answer:

D

Explanation:

D) The overall work done by gravity is zero  

This statement is correct .

If m be the mass of each of the children and h be the height of tower

work done by gravity on the boys in going up = - mgh

it is so because force applied by gravity = mg downwards and displacement

is upwards

work done will be negative = - mgh

Work done by gravity on boys when they come down = + mgh because both force and displacement are downwards .

Hence total work done = - mgh + mgh = 0.

The children will have same kinetic energy as the inclined surface is friction-less so no energy will be dissipated hence addition of energy to boys in both the cases will be same.

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When sugar is poured from the box into the sugar bowl, the rubbing of sugar grains creates a static electric charge that repels
Afina-wow [57]

Answer:

2.6×10⁻³ N

Explanation:

From coulomb's law,

F = kq'q/r²................ Equation 1

Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.

From the question,

since q' = q

Then,

F = kq²/r²..................... Equation 2

Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,

Constant: k = 9×10⁹ Nm²/kg².

Substitute into equation 2

F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²

F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)

F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)

F = 2.6×10⁻³ N.

3 0
3 years ago
A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider
julia-pushkina [17]

There's not enough information to find an answer.

I think the idea here is that in descending (416 - 278) = 138 meters,
the glider gives up some gravitational potential energy, which
becomes kinetic energy at the lower altitude.  This is all well and
good, but we can't calculate the difference in potential energy
without knowing the mass of the glider.

3 0
4 years ago
Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.
Gelneren [198K]

Answer:

2.99\cdot 10^{11}C

Explanation:

The mass of one electron is

m_e = 9.11\cdot 10^{-31}kg

So the number of electrons contained in M=1.7 kg of mass is

N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}

The charge of one electron is

e=1.60\cdot 10^{-19} C

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C

8 0
3 years ago
A bicyclist is in a 50-km race. She says she had an average velocity of 35.
zlopas [31]
The correct answer to this is (A. Units Only).

It shows that there is a velocity of 35, but the units are missing.
8 0
4 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
4 years ago
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