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Sedbober [7]
3 years ago
10

7. A Candle is placed 4 cm in front of a convex lens. The image of the candle is focused on a sheet of paper that is exactly 10

cm behind the lens
What is the Magnification of the image​
Physics
1 answer:
velikii [3]3 years ago
3 0

Answer:

Magnification m is 2.5cm

Explanation:

This problem bothers on lenses

Given data

Object distance u= 4cm

Height of object v= 10 cm

The problem is quite straightforward seeing that object and image distances are given

We know that the magnification

m= size of image/size of object

m= v/u= 10/4 = 2. 5cm

What is a convex lens

A convex lens is a type of lens that has the centre thicker than the ends, the thickness at the centre makes the lens surface to curve outward. Convex lens causes close light rays of light to converge at a point after refraction.

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If the car goes exits a freeway and goes from 65<br> mph to 35 mph is it accelerating?
Zolol [24]

Answer:

No, the car is decelerating  

Explanation:

No the car is decelerating if it exits a freeway and goes from 65

mph to 35 mph since the change in velocity is negative.

change in velocity = final - initial

change in velocity  = 35 - 65

change in velocity = -30mph

Since the change in velocity is negative, hence the car is decelerating. Deceleration is a negative acceleration

8 0
3 years ago
What is the accletation of a 1500kg with a net force of 7500 N
matrenka [14]
Acceleration formulae is:
a=Fnet/mass
According to the question
a=7500N/1500kg
a=5m/s sq.
3 0
3 years ago
Two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of
Arte-miy333 [17]

Answer:

c. because A will land first becuase its heavier

Explanation:

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2 years ago
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8 0
3 years ago
A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc
liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s,    a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

           v₀ₓ =  10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

           v_y = v_{oy} - g t

           v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)

at the highest point of the trajectory the vertical speed must be zero

           v_y = 0

therefore the velocity of the body is

          v = (10 i ^ + 0j ^) m / s

the acceleration is

          a = (0 i ^ - g j⁾

          a = (0i ^ - 9.8 j ^) m / s²

5 0
2 years ago
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