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aivan3 [116]
2 years ago
5

1) A marble, rolling with speed 20 cm/s, rolls off the edge of a table that is 80 cm high.

Physics
1 answer:
Schach [20]2 years ago
4 0
1) A marble, rolling with speed 20 cm/s, rolls off the edge of a table that is 80 cm high.
(a) How long does it take to drop to the floor?
(b) How far, horizontally, from the table edge does the marble strike the floor?
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Someone help please i need to finish this
Tom [10]

Answer:

4th answer

Explanation:

The gradient of a distance-time graph gives the speed.

gradient = distance / time = speed

Here, the gradient is a constant till 30s. So it has travelled at a constant speed. It means it had not accelarated till 30s. and has stopped moving at 30s.

6 0
2 years ago
A grain silo is shown below:grain silo formed by cylinder with radius 5 feet and height 175 feet and a half sphere on the topwha
nekit [7.7K]
Find the volume of the bottom and top separately and then add them.
Cylinder volume is the area of the bottom times the height
(22/7)(5^2)•175=13750 ft^3

The volume of a sphere is
V=(4/3)(22/7)r^3
where r is the radius. Here that's also 5 since it fits on the cylinder.
Also we only want half the sphere so use
V=(2/3)(22/7)•5^3=261.9 ft^3
Which we round upto 262.
Now add the parts together
13750+262=14,012 ft^3
6 0
3 years ago
propane, the gas used in barbeque grills, is made of carbon and hydrogen. Will the atoms that make up propane form covalent bond
nirvana33 [79]
<span>Hydrocarbons are molecules that contain only carbon and hydrogen.</span> Due to carbon's unique bonding patterns, hydrocarbons can have single, double, or triple bonds between the carbon atoms. The names of hydrocarbons with single bonds end in "-ane," those with double bonds end in "-ene," and those with triple bonds end in "-yne". The bonding of hydrocarbons allows them to form rings or chains.
8 0
3 years ago
Read 2 more answers
The atomic mass of an element is
inessss [21]

Here are the answers to the question. Make sure to give a valid reason, please.

A. the sum of the protons and neutrons in one atom of the element.

B. a ratio based on the mass of a carbon-12 atom.

C. a weighted average of the masses of an element's isotopes.

D. twice the number of protons in one atom of the element.

6 0
3 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
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