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Sladkaya [172]
3 years ago
14

Where should an object be placed in front of a concave mirror’s principal axis to form an image that is real, inverted, larger t

han the object, and farther from the mirror than the object is?
Physics
2 answers:
Charra [1.4K]3 years ago
7 0
If the object is kept in between the principle axis and the focus but some what nearer to the focus then we will get the enlarge,erect,and real image.

Alja [10]3 years ago
6 0

Answer:

At focus.

Explanation:

When the object is placed at the focus of a concave mirror, the image will form at infinity. The formed image is real and inverted and its size is highly enlarged.

When the object is placed between the pole and the focus, the image will formed behind the mirror. The formed image is virtual and erect and its size is enlarged. This is the only case when the formed image is virtual and erect.

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it is about 384,750 kilometers from earth to the moon. it took the apollo astronauts about 2 days and 19.5 hours to fly to the m
Jet001 [13]
We know, speed = Distance / Time
d = 384,750 Km
t = 2 days, 19.5 hours = 48+19.5 = 67.5 hour

Substitute their values, 
s = 384,750 / 67.5
s = 5700 Km/h

In short, Your Answer would be 5700 Km/h

Hope this helps!
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3 years ago
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Jobisdone [24]

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Directions: Fill in the blank with the correct word. Choose from the list of possible answers in the word bank below:
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1).atoms (3). mixture. (5). Element

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8 0
3 years ago
Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at r
Deffense [45]

Answer:

m=57.65 kg

Explanation:

Given Data

Ricardo mass m₁=80 kg

Canoe mass m₂=30 kg

Canoe Length L= 3 m

Canoe moves x=40 cm

When Canoe was at rest the net total torque is zero.

Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as

m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)

We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.

so

2x=40

x=40/2

x=20 cm

Substitute in the above equation we get

m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg

4 0
3 years ago
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