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lana66690 [7]
1 year ago
11

Determine the image distance and image height for a 5.00 cm tall object placed 30.0 cm from a double convex lens with a focal le

ngth of 15.0 cm.
Physics
1 answer:
vlabodo [156]1 year ago
8 0

Answer:

The image distance is 30 cm

image height = - 5 cm

Explanation:

The formula for calculating the image distance is expressed as

1/f = 1/u + 1/v

where

f is the focal length

u is the object distance

v is the image distance

From the information given,

u = 30

f = 15

By substituting these values into the formula,

1/15 = 1/30 + 1/v

1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30

Taking the reciprocal of both sides,

v = 30

The image distance is 30 cm

magnification = image height/object height = - v/u

Given that object height = 5 cm, then

image height/5 = - 30/30 = - 1

image height = - 5 * 1

image height = - 5 cm

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Give one example of a question that science would not be able to test. Then, explain how it could be changed into a testable que
AnnZ [28]
Who was the first president?

It could be changed by asking
“Who was the first president and what was there scientific accomplishment?”
3 0
3 years ago
An arrow is shot at a target 20 m away. The arrow is shot with a horizontal velocity of 80 m/s.
Verdich [7]

(1) The time of motion of the arrow is 0.25 s.

(2) The vertical height dropped by the arrow as it approaches the target is 0.31 m.

The given parameters:

  • <em>Horizontal distance of the arrow, X = 20 m</em>
  • <em>Horizontal speed of the arrow, v = 80 m/s</em>

<em />

The time of motion of the arrow is calculated as follows;

t = \frac{X}{v} \\\\t = \frac{20 }{80} \\\\t  = 0.25 \ s

The vertical height dropped by the arrow as it approaches the target is calculated as follows;

h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 \ + \ \frac{1}{2} \times 9.8 \times 0.25^2\\\\h =0.31 \ m

Learn more about time of motion of projectile here:  brainly.com/question/1912408

4 0
2 years ago
What is the difference between charging by contact and charging by induction in terms of electron transfer.
Veronika [31]

Answer:

the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch

Explanation:

There are three methods of charging an object:

- Charging by friction: this is done by rubbing an object against another object. An example is when a plastic rod is rubbed with a wool cloth. When this is done, electrons are transferred from the wool to the rod, so both objects remain charged at the end of the process

- Charging by contact: this is done by putting in contact a charged object with a neutral, conducting object. In this case, the charges are transferred from the charged object to the neutral object; at the end of the process, the neutral object will also have a net electric charge, so it will be also charged.

- Charging by induction: in this case, we take a charged object, and a neutral object, and we place the two objects close to each other, but without touching. Let's assume that the charged object is negatively charged: in this case, the positive charges in the neutral object are attracted towards the negative charges of the charged object, while the negative charges of the neutral object are repelled away. As a result, the positive and negative charges in the neutral object split apart. If the object is connected to the ground, then negative charges move away, so the neutral object will remain positively charged.

Therefore, the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch.

5 0
3 years ago
Suatu sistem gas berada didalam ruang yang fleksibel. Pada awalnya gas berada pada kondisi P1 = 1,5 × 10 pangkat 5 N/m2, T1 = 27
enot [183]

Answer:

16.00L

Explanation:

First you calculate the number of moles in the system:

PV=nRT\\\\n=\frac{PV}{RT}\\\\n=\frac{(1.5*10^5N/m^2)(12L)}{(0.082L.atm/mol.K)(300.15K)}=73134.16\ mol

To find the new volume of the system you use the following formula for an isobaric procedure:

T_2-T_1=\frac{P}{nR}(V_2-V_1)\\\\V_2=\frac{nR}{P}(T_2-T_1)+V_1\\\\V_2=\frac{(73134.16\ mol)(0.082L.atm/mol.K)}{1.5*10^5N/m^2}(400.15-300.15)K+12L\\\\V_2=16.00L

hence, the new volume is 16.00L

6 0
4 years ago
The body of mass 1 kg is kept at rest. A constant force of 6.0 N acting on it the time taken by the body to move through a dista
HACTEHA [7]

Time taken by the body to move through a distance of 12 m is 2 s

Explanation:

using the equation for force

F= m a

m= mass= 1 kg

F= force= 6 N

a= acceleration=

6= 1 a

a= 6 m/s²

Now using the kinematic equation,

d= Vt + \frac{1}{2} at^2

d= distance= 12 m

V= initial velocity=0 as the object is at rest initially

12= 0(t)+ 1/2 (6) t²

t²=4

t= 2 s

8 0
3 years ago
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