Answer:
Pinhole Projection
Explanation:
You will get a great view if you project a sunlight coming thrugh a small hole onto a viewing card enclosed in a long box. The smaller the pinhole, the sharper the image
Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is
<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²
where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².
The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is
<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²
<em>F</em> ≈ 9.81 N
Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.
This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of
<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²
i.e. 1/4 of the weight on Earth, which would be about 2.45 N.
Iodine-131 has a half life of 8 days, so half of it is gone every 8 days.
10 grams of iodine-131 is left for 24 days.
At 8 days: 10/2=5 grams left
At 16 days: 5/2=2.5 grams left
At 24 days: 2.5/2=1.25 grams left.
**
Your mistake is that you stopped at 16 days.
Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
Answer:I need the ans to this one too
Explanation: