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netineya [11]
3 years ago
6

An object is placed at a distance of 60 cm from a concave lens of focal length 30cm.find the position and nature of the image?

Physics
1 answer:
ladessa [460]3 years ago
5 0

Answer:

using the lens formula: 1/f = 1/u + 1/v

focal length f = -30 (negative because it is concave lens)

object distance u = 60

image distance v= unknown

1/-30 = 1/60 + 1/v

v = -20

So, the image is 20cm from lens (on the same side along with the object), and it is virtual (because of negative sign) and erect (concave lens must produce erect images).

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Answer:

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Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb  

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

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Initial momentum of car B=mv where m is mass and v is velocity in ft/s  

46 mph*1.46667=67.4666668 ft/s

Momentum_B=1125*67.4666668 ft/s

Initial momentum of car A is given by

Momentum_A=1515v_a where v_a is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

1515v_a-(1125*67.4666668 ft/s)

The common velocity is represented as v_c hence after collision, the final momentum is

Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

1515v_a-(1125*67.4666668 ft/s)= 2640v_c

The acceleration of two cars a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}

From kinematic equation

v^{2}=u^{2}+2as hence

v^{2}-u^{2}=2as

0^{2}-(v_c)^{2}=2*-24.1275*19.5

v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s

Substituting the value of v_c in equation 1515v_a-(1125*67.4666668 ft/s)= 2640v_c

1515v_a-(1125*67.4666668 ft/s)= 2640*30.67

1515v_a=(1125*67.4666668 ft/s)+2640*30.67

v_a=\frac {156868.8}{1515}=103.5438 ft/s

\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph

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