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3 years ago

Which model below describes the processes of Gamma radiation? A B. C. D.

Physics

1 answer:

S_A_V [24]3 years ago

5 0

Answer:

Explanation:

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Hello, I want to ask. . anyone knows the answer.

sasho [114]

Answer:

Explanation:

Please give brainliest.

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3 years ago

________________ is the tendency of an object to resist a change in its motion.

Sliva [168]

Inertia is the tendency of an object to resist a change in its motion.

7 0

3 years ago

1. Our entire solar system orbits around the center of the about once every 230 million years. 2. The Milky Way and Andromeda ga

bekas [8.4K]

Answer: sorry mate but what's the question here tho?

Explanation:

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3 years ago

A brick is dropped from a high scaffold. <br> a. How far does the brick fall during this time?

Alecsey [184]

Answer:

a: after 1 seconds it will have fallen 0.2452

after 2 seconds it will have fallen 0.981

after 3 seconds it will have fallen 2.2072

after 4 seconds it will have fallen 3.924

Explanation:

the formula for acceleration due to gravity is (ignoring friction I think)

g = G*M/R^2

earths gravitational constant is about 9.807

g = 9.807*M/R^2

The average weight of a brick is 5 pounds and I'm going to say it's 10 feet off the ground.

g = 9.807*5/10^2. g = 0.4905 so every second the brick will go 0.4905 fps faster. (fps means feet per second.)

after 1 seconds it will have fallen 0.2452

after 2 seconds it will have fallen 0.981

after 3 seconds it will have fallen 2.2072

after 4 seconds it will have fallen 3.924

6 0

3 years ago

A firm receives an order for a square-base rectangular storage container with a lid. The container has a volume of 20 cubic mete

Stels [109]

Answer:$ 506.05

Explanation:

Given

volume of container $=20 m^3$

Let L be the length of square-base and h be the height of Rectangular box

Cost of base $=20 \$/m^2$

Cost of side and lid $=10 \$ /m^2$

Cost of base $c_1=L^2\times 20$

$h=\frac{20}{L^2}$

cost of lid and side $c_2=10\times 4L\cdot h+10\times L^2$

Total cost $C=c_1+c_2$

$C=20L^2+10L^2+40L\cdot h$

$C=30L^2+\frac{800}{L}$

differentiate C w.r.t to L to get minimum cost

$\frac{\mathrm{d} C}{\mathrm{d} L}=60L-\frac{800}{L^2}=0$

$L^3=\frac{80}{6}$

$L=2.37 m$

thus $h=\frac{20}{5.613}=3.56 m$

Thus Lowest cost is $C=30\times 5.617+\frac{800}{2.37}=$ 506.05$

4 0

3 years ago