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stira [4]
3 years ago
13

7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent

Physics
1 answer:
Rudiy273 years ago
5 0

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm  (Approx)

Explanation:

Given:

Object distance u = 12 cm

a) Focal length = 20 cm

b) Focal length = 16 cm

Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

v = 7.5 cm

Lens at a distance = 7.5 cm

b.  1/v = 1/u + 1/f

1/v = 1/16 + 1/12

v = 6.86 cm  (Approx)

Lens at a distance = 6.86 cm  (Approx)

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Answer:

Explanation:

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c )

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d )

Power in primary = power in secondary = 16.67 W

e ) current drawn by ac line ( primary )

Volt ( primary ) x current ( primary ) = power in primary

= 16.67

current ( primary )

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3 years ago
What type of system would allow light to enter and exit, but would keep any
andrew11 [14]
An isolated system , it does not allow any matter or energy to be exchanged
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3 years ago
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A ski lift carries a 75.0-kg skier at 3.00 m/s for 1.50 min along a cable that is inclined at an angle of 40.0° above the horizo
Nookie1986 [14]

Answer

given,

mass of the ski = 75 Kg

speed of the skier, v = 3 m/s

time = 1.50 min = 90 s

angle of inclination, θ = 40°

distance = s x t

              = 3 x 90 = 270 m

a) W = F. d cos θ

   W = mg. d cos θ

   W = 75 x 9.8 x 270 x cos 40°

    W = 152021.52 J

work is done by the ski lift is equal to  152021.52 J

b) Power extended by the ski

 Power = \dfrac{Work\ done}{time}

Power = \dfrac{152021.52}{90}

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power is expended by the ski lift is equal to 1689.13 W.

3 0
3 years ago
If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val
baherus [9]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}

\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \  body \ (v) = 20 \ m/s}

\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \  body \ (u) = 0}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \  body \ ( F)}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

☯ <u>Using 1st equation of motion </u>

\\

\dashrightarrow\:\: \sf{v = u + at}

\\

\dashrightarrow\:\: \sf{20 = 0 + a(4)}

\\

\dashrightarrow\:\: \sf{20 = 4a}

\\

\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}

\\

\dashrightarrow\:\: \sf{a = 5}

\\

☯ <u>Now, Finding the force exerted </u>

\\

\dashrightarrow\:\: \sf{F = ma}

\\

\dashrightarrow\:\: \sf{F = 40 \times 5}

\\

\dashrightarrow\:\: \sf{F = 200 \ N}

\\

☯ <u>Hence</u>, \\

\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}

8 0
3 years ago
As Earth moves around the Sun, its _____ is approximately perpendicular to the force of gravity exerted by the Sun.
Bogdan [553]

SORRY i forgot but i think its A because the sun has a really strong gravitational pull


sorry and hope it helps

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3 years ago
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