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Daniel [21]
4 years ago
15

18. If the distance from a converging lens to the object is less than the focal length of the lens, the image will be _______ th

e object. A. virtual, upright, and smaller than B. real, inverted, and larger than C. virtual, upright, and larger than D. real, inverted, and smaller than
Physics
1 answer:
skad [1K]4 years ago
5 0
Here's a quick way to find out. Pick up your glasses, bifocals work best, and find the focal length with a flashlight against a book. If I remember right, the object should be magnified and upside down. So, A.
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Why are there the 3 kinds of AC power?
Roman55 [17]
There are three major kinds of power supplies: unregulated (also called brute force), linear regulated, and switching. The fourth type of power supply circuit called the ripple-regulated, is a hybrid between the “brute force” and “switching” designs, and merits a subsection to itself.
6 0
3 years ago
1. Find the magnitude of the gravitational force (in N) between a planet with mass 8.00 ✕ 1024 kg and its moon, with mass 2.75 ✕
MrRissso [65]

Answer:

1.358 10e8 have good day please mark brainliest

8 0
3 years ago
This question is divided into two parts. This is part (a) of the question. A proton accelerates from rest in a uniform electric
grin007 [14]

Answer:

The acceleration of proton is 5.56 x 10^10 m/s^2 .

Explanation:

initial velocity, u = 0

Electric field, E = 580 N/C

final speed, v = 10^6 m/s

(a) Let the acceleration is a.

According to the Newton's second law

F = m a = q E

where, q is the charge of proton and m is the mass.

a= \frac{q E}{m}\\\\a = \frac{1.6\times10^{-19}\times 580}{1.67\times 10^{-27}}\\\\a= 5.56\times 10^{10} m/s^2

4 0
3 years ago
What happens to parallel light rays that strike a concave lens?
zhenek [66]
Due to the shape of the lens , parallel rays will be deviated
5 0
3 years ago
Read 2 more answers
An object experiences an acceleration of 8.5 m/s^2 over a distance of 300 m. After that acceleration it has a velocity of 400 m/
Snezhnost [94]

Answer:

393.6m/s

Explanation:

Given parameters:

Acceleration  = 8.5m/s²

Distance  = 300m

Final velocity  = 400m/s

Unknown:

Initial velocity  = ?

Solution:

To solve this problem, we use the expression below;

             v² = u²  + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

      So;

               v²  - 2as = u²

        u²   = v²   - 2as

        u²  = 400²   - (2 x 8.5 x 300)  

         u   = 393.6m/s

7 0
3 years ago
Read 2 more answers
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