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9 days ago

The left end of a long glass rod 6.00 cm in diameter has a convex hemispherical surface 3.00 cm in radius. The refractive index

of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end:
(a) infinitely far,
(b) 12.0 cm;
(c) 2.00 cm.

Physics

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A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Softa [3030]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Consider the following:

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field measured at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

Thus,

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now, by applying integration to the equation above

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

3 months ago

A submarine dives from rest a 100-m distance beneath the surface of an ocean. Initially, the submarine moves at a constant rate

Keith_Richards [3271]

a. Time = 16.11 s b. Gauge Pressure = 1009400 Pa = 1 MPa c. Absolute Pressure = 1110725 Pa + 1.11 MPa d. Force = 2.22 MN

8 0

1 month ago

The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the

kicyunya [3294]

Answer:

Explanation:

An image of the bond resulting from the search is attached.

Consider the force directed towards thymine as negative.

For the O-H-N combination:

The resulting force from this combination is:

$F=-F_{OH}+F_{ON}\\\\=\frac{Ke^2}{r^2}+\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[(0.280-0.110)\times 10^{-9}m]^2}+\frac{1}{0.280\times 10^{-9}m)^2}]\\\\=-5.03354\times 10^{-9}N$

In the case of the N-H-N combination:

The total force acting from this combination is:

$F'=F_{NN}-F_{HN}\\\\=\frac{Ke^2}{r^2}-\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[0.300\times 10^{-9}m]^2}-\frac{1}{((0300-0..110)\times 10^{-9})m)^2}]\\\\=-3.822\times 10^{-9}N$

The force that thymine applies on adenine is:

$F_{net}=F+F'\\\\=-5.03354\times 10^{-9}N-3.822\times 10^{-9}N\\\\=-8.8558\times 10^{-9}N$

When rounded to three significant figures, the net force is $8.86\times 10^{-9}N$

The negative value indicates that the force is attractive, as it is aimed towards thymi.

The force acting on the electron due to the proton is:

$F=\frac{Ke^2}{r^2}\\\\=\frac{(9.0\times 10^9Nm^2/C^2)(1.6\times 10^{-9}C)^2}{(5.29\times 10^{-11}m)^2}\\\\=8.233\times 10^{-8}N$

Since the electron and proton carry opposite charges, the force on the electron points towards the proton.

The ratio of the above forces is:

$\frac{F}{F_{net}}=\frac{8.233\times 10^{-8}N}{8.233\times 10^{-8}N}\\\\=9.3$

Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.

5 0

3 months ago

A 1 kg ball moving horizontally to the right at 3 m/s strikes a wall and rebound, moving horizontally to the left at the same ve

Maru [3345]

Answer:

The change in linear momentum of the ball amounts to 6 kg-m/s.

Explanation:

Provided that,

Ball mass, m = 1 kg

Initial ball speed, u = 3 m/s

Upon striking a wall, it rebounds, moving horizontally to the opposite side at the same speed of 3 m/s, thus v = -3 m/s

The change in linear momentum is calculated as follows:

$p=m(v-u)\\\\p=1\ kg\times (-3-3)\\\\p=-6\ kg-m/s$

Consequently, the magnitude of the change in linear momentum of the ball is 6 kg-m/s.

3 0

3 months ago