Relative to the distance of an object in front of a plane mirror, how far behind the mirror is the image?

S Problem Set<br> 2.) 6.4 x 109 nm to cm

anyanavicka [17]

Answer:

$6.4\cdot 10^2 cm$

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

$1 nm = 10^{-9} m$

So we have:

$6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m$

Now we can convert from metres to centimetres, keeping in mind that

$1 m = 10^2 cm$

So, we find:

$6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm$

8 0

3 years ago

An amusement park ride consists of a rotating circular platform 10.9 m in diameter from which 10 kg seats are suspended at the e

Darya [45]

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4 0

3 years ago

two horses pull against a rope with forces of 100 newtons in opposite directions. this is an example of

statuscvo [17]

It is an example of balanced force.

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4 0

3 years ago

Read 2 more answers

Halley's comet orbits the sun roughly once every 76 years. It comes very close to the surface of the Sun on its closest approach

Licemer1 [7]

Answer:

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

3 0

3 years ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

Drupady [299]

Answer:

$4.5\times 10^{-5} T$

Explanation:

We are given that

Current in wire=40 A

Magnetic field= $B_1=3.5\times 10^{-5}$ T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

$B_{wire}=B_2=\frac{\mu_0I}{2\pi R}$

We have R=29 cm= $\frac{29}{100}=0.29 m$

1 m=100 cm

Substitute the values in the given formula

$B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T$

The resultant magnetic field is given by

$B=\sqrt{B^2_1+B^2_2}$

Substitute the values then we get

$B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}$

$B=4.5\times 10^{-5} T$

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire= $4.5\times 10^{-5} T$

Hence, the resultant magnitude of the magnetic field 29 cm above the wire= $4.5\times 10^{-5} T$

7 0

3 years ago