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Mamont248 [21]
3 years ago
5

How do you think the amount of a material affects its tendency to sink or float?

Physics
1 answer:
Viktor [21]3 years ago
6 0

Answer:

Things float when they are positively buoyant, or less dense than the fluid in which they are sitting. This does not mean that an object has to be lighter than the fluid, as in the case of a boat; objects just need to have a greater ratio of empty space to mass than the fluid.

Explanation:

pls mark as brainliest

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plastic bags have replaced for the packing of the most materials. give these advantages and two disadvantages of using plastic o
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It is waterproof and more durable
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A Ferris wheel has diameter of 10 m. It rotates at a uniform rate and makes one revolution in 8.0 seconds. A person weighing 670
Nikolay [14]

Answer:  459.14 N

Explanation:

from the question, we have

diameter = 10 m

radius (r)  = 5 m

weight (Fw) = 670 N

time (t) = 8 seconds

Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path, therefore we apply the equation below

∑ F = F c =  F w − Fn ..............equation 1

Fn = Fw − Fc = mg − (mv^2 / r) ...................equation 2

substituting the value of v as (2πr / T) we now have

Fn = mg − (m(2πr / T )^2) / r

Fn= mg − (4(π^2)mr / T^2)   ..........equation 3

Fw (mass of the person) = mg

therefore m = Fw / g

                m = 670 / 9.8 = 68.367 kg

now substituting  our values into equation 3

Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)

Fn = 670 - 210.86

Fn = 459.14 N

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3 years ago
Identify the region in the image where lighter elements, such as hydrogen or helium, are more likely to be found in a liquid or
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These graphs show measurements of the density of air as different sound waves pass a single point.
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A fireworks shell is accelerated from rest to a velocity of 68.0 m/s over a distance of 0.230 m.
Ugo [173]

Answer:

(A) 10052.2 m/s²

(B)  0.00678 seconds

Explanation:

From the question,

(A) Applying

V² = U²+2as..................... Equation 1

Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.

make a the subject of the  equation

a = (V²-U²)/2s........................ Equation 2

Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m

Substitute these values into equation 2

a = (68²-0²)/(2×0.230)

a = 10052.2 m/s²

(B) Using,

a = (V-U)/t......................... Equation 3

Where t= time.

make t the subject of the equation

t = (V-U)/a......................... Equation 4

Given: V = 68 m/s, U = 0 m/s, a = 10052.2

Substitute into equation 4

t = (68-0)/10052.2

t = 0.00678 seconds

5 0
3 years ago
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