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Virty [35]
3 years ago
6

A girl stands 5m away from a large plane mirror. How far must she walk to be 2m away from her image?

Physics
1 answer:
lord [1]3 years ago
8 0

Answer:

The girl needs to walk 4m closer to the mirror if she must be 2 m away from her image.

Explanation:

Images formed by plane mirrors is due to reflection of light Plane mirrors form only virtual or imaginary images. The image formed by a plane mirror is equidistant from the mirror as the object is from the mirror. This means that the distance of the girl from the mirror which is 5m is equal to the distance of her image from the mirror.

The distance of the girl from her image when she stands 5 m from the mirror is therefore, 5m + 5m = 10 m

In order for her to be 2 m away from her image, she needs to sand at a distance half her distance from her image, that is 1/2 * 2m = 1m

Therefore, the girl needs to walk 5m - 1m = 4m closer to the mirror if she has to be 2 m away from her image.

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During anomalous expansion of water ,heat transfer is limited to conduction and radiation only ,explain.​
aalyn [17]

Heat transfer is limited to conduction and radiation only in anomalous expansion of water simply because of the temperature at which the expansion occurs and density

<h3>What is anomalous expansion of water?</h3>

Anomalous expansion of water is a property of water in which water expands instead of contracting.

  • Anomalous expansion of water makes water less dense.
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Learn more about properties of water:

brainly.com/question/18681949

4 0
2 years ago
An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis
Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (\Delta K) and part is lost because of the air resistance (E_{lost}).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

where

v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

8 0
3 years ago
Durante uma corrida de carros, o piloto da escuderia ganhadora percorreu a primeira volta da pista com velocidade media de 310 k
lesya692 [45]
A velocidade mínima é 0. A velocidade máxima é inferior ou igual a 620 km/h.
<span>buena suerte mi amigo</span>
3 0
3 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
4 years ago
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