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telo118 [61]
4 years ago
12

Imagine tying a string to a ball and twirling it around you. How is this similar to the moon orbiting Earth? In this example, wh

at is providing the constantly changing, inward force?
Physics
1 answer:
leonid [27]4 years ago
5 0
The string kind of acts like gravity 
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Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?
Schach [20]
Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.
6 0
3 years ago
True or false question please help me out
Len [333]

Answer:

thats true

Explanation:

7 0
2 years ago
What is used to block the UV light during screen development process to create a stencil in the emulsion
coldgirl [10]

Answer:

Invisible UV energy reacts with emulsion sensitizer and hardens the stencil so it won't dissolve with water and rinse down the drain

Explanation:

4 0
2 years ago
When driving straight down the highway at a constant velocity you have to give the engine a little gas (which means an added ext
Alona [7]

Answer:

Explanation:

This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.

7 0
3 years ago
A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is
Rudik [331]

Answer:

a=16.2m/s^{2}

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\

When fixed,F=o

Hence

masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}

The value of the acceleration is 16.2m/s^2

7 0
3 years ago
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