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Misha Larkins [42]
3 years ago
15

Which of the following represents a concave lens? A. -di B. +di C. -f D. +f

Physics
2 answers:
seraphim [82]3 years ago
6 0

for a concave lens it should be -f

Nina [5.8K]3 years ago
4 0

Answer:

The answer is option D. +f

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An airplane capable of an airspeed of 100 km/hr is 60 km off the coast above the sea. If the wind is blowing from the coast out
Nady [450]

To solve this problem we will apply the concepts related to relative speed. We will obtain it from the deduction made on the aircraft as a speed of the two components that act on it. Through the kinematic equations of motion, we can then calculate the time required.

The airspeed of airplane is 100km/h  while the wind is blowing from the coast out to sea at 40km/h. Wind is blowing from the coast out to sea means that it opposes the airspeed. Therefore, resultant relative speed of airplane is

v_r = 100-40=60km/h

Total distance is 60km then with this net velocity we have that the required time is

v = \frac{x}{t} \rightarrow t = \frac{x}{v}

Where,

x = Displacement

t = Time

v = Velocity

Replacing,

t = \frac{60km}{60km/h} = 1hour

t = 60 minutes

Therefore the time taken by the plane to reach the shore is 60 minutes

6 0
3 years ago
What will terminal velocity look like on a velocity v. time graph?
mrs_skeptik [129]
The horizontal is s or displacement
The vertical is t or time
3 0
2 years ago
What is ligma please answer asap
Goshia [24]

ligma nuts is it the answer bro ;)



6 0
3 years ago
Read 2 more answers
4 points
zubka84 [21]

(a) 25lx

(b) 11.11lx

<u>Explanation:</u>

Illuminance is inversely proportional to the square of the distance.

So,

I = k\frac{1}{r^2}

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

I_2 = k\frac{1}{(2r)^2}

Dividing both the equation we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4}  = 25lx

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

I_2 = k\frac{1}{(3r)^2}

Dividing equation 1 and 3 we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9}  = 11.11lx

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0
3 years ago
Increasing the two objects will cause the gravitational force between the objects to decrease.
ELEN [110]
This statement is false. Increasing the two objects' mass (I'm guessing) will actually increase their gravitational force. This is because of the equation:

F_g =  \frac{Gm_1m_2}{d^2}

If the distance was increased, then the statement would be true, but since you are increasing mass, which is proportional to the Force of Gravity, you are in fact, increasing the gravitational force between the two objects.
4 0
3 years ago
Read 2 more answers
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