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xz_007 [3.2K]
3 years ago
10

12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and

size of the image. (3)​
Physics
1 answer:
Nimfa-mama [501]3 years ago
6 0

Answer:

I think 9.5

Explanation:

............

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Why is it dangerous to connect a battery to itself. what is the term used to describe that
Yuki888 [10]
In the circuit outside of the battery the electrons have to expend all of their energy on the internal resistance of the battery which causes heating
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Continental crust is primarily witch kind of rock
kotegsom [21]
The correct answer is granite oceanic. Continental crust is primarily a granite oceanic. Continental crust is the layers of sedimentary, igneous, and metamorphic rocks that form the area of the shallow seabed that is near the shore. 
7 0
3 years ago
If your car runs out of gas and you must push it 250 m to the nearest gas station. How much work is done on the car if it is pus
tankabanditka [31]

Answer:

1750 Joules.

Explanation:

Work done = force * distance

= 7 * 250

= 1750 Joules.

6 0
2 years ago
A car with a mass of 850kg is moving at a speed of 72km/h when colliding with a concrete wall until it stops. After the collisio
Sergeeva-Olga [200]

Answer:

Explanation:

The vehicle is experiencing a large force created by the concrete wall.

Equation

vf^2 = vi^2 + 2*a * d

Givens

vf = 0   The car eventually does stop.

vi = 72 km/hr * [ 1000 m/  km] * [1 hour / 3600 seconds]

vi = 20 meters / second

a = ?

m = 850 kg

Solution

vf^2 = vi^2 + 2a*d

0 = 20 m/s + 2* 2 *a

-20 m/s = 4a

-20/4 =   a

a = - 5 m/s^2   The minus sign tells you the vehicle is slowing down. It sure should be.

Force = m * a

F = - 850 * (-5)

F =  - 4250 N

The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.

8 0
3 years ago
A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in
dexar [7]

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860  m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}

Put all the values,

t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s

The velocity of the ball is :

v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s

The kinetic energy of the ball is :

K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J

So, the required kinetic energy is 0.076 J.

6 0
3 years ago
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