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Nostrana [21]
3 years ago
10

A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change

the period to 2.00 s
Physics
1 answer:
Daniel [21]3 years ago
7 0

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

T = 2\pi \sqrt{\frac{m}{k} }

Where K indicates spring constant

m indicates mass

For the new time period

T^' = 2\pi \sqrt{\frac{m'}{k} }

Now, we will take 2 ratios of the time period

\frac{T}{T'} = \sqrt{\frac{m}{m'} }

\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }

0.5625 = \sqrt{\frac{0.500}{m'} }

m' = \frac{0.500}{0.5625}

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

or

= 389 kg

Therefore for computing the mass we simply applied the above formula.

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Solutions with ions that react with acids or bases to lessen their effects are fubrefs
Softa [21]

Answer:

Buffers

Explanation:

A buffer solution is a solution containing weak acids and their salts or weak bases and their salts.

A buffer solution is an equilibrium system that resists changes in pH or pOH when a small amount of an acid or base is added hence it is a solution of fairly constant pH value.

4 0
3 years ago
A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.
gtnhenbr [62]

Answer:

(a)  vmax = 0.34m/s

(b)  v = 0.13m/s

(c)  v = 0.31m/s

(d)  x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

v_{max}=\omega A       (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

\omega=\sqrt{\frac{k}{m}}           (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

x=Acos(\omega t)

You solve the previous equation for t:

t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s

With this value of t, you can calculate the speed of the object with the following formula:

v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s

The position will be:

x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0
4 years ago
Absolute differences between each observed mass and the accepted value,divided by the accepted value, multiplied by 100%
timama [110]

Answer:

The percentage error    e% = Δm/m 100    

Explanation:

The absolute error is the assessment with which measurements are made, it is given by the minimum reading value of a given instrument.

The accepted value is the value measured in an instrument. Or the one calculated by an experiment.

The percentage error is the ratio of the absolute error between the value of the average per 100

      e% = Δm/m 100

3 0
3 years ago
How much work does a 120 watt motor perform for 300 seconds?
MrRa [10]

Answer:

36,000j

Explanation:

W= P•t

120•300= 36000

I checked my answer with P= W/t and got the right answer

hope this helps

6 0
3 years ago
How are both chemistry and physics related to geoscience?
Yanka [14]

Answer:

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There are both reductionist and holistic approaches to Earth sciences. It is also the study of Earth and its neighbors in space. Some Earth scientists use their knowledge of the planet to locate and develop energy and mineral resources. Others study the impact of human activity on Earth's environment, and design methods to protect the planet. Some use their knowledge about earth processes such as volcanoes, earthquakes, and hurricanes to plan communities that will not expose people to these dangerous events.

The Earth sciences can include the study of geology, the lithosphere, and the large-scale structure of the Earth's interior, as well as the atmosphere, hydrosphere, and biosphere. Typically, Earth scientists use tools from geology, chronology, physics, chemistry, geography, biology, and mathematics to build a quantitative understanding of how the Earth works and evolves. Earth science affects our everyday lives. For example, meteorologists study the weather and watch for dangerous storms. Hydrologists study water and warn of floods. Seismologists study earthquakes and try to understand where they will strike. Geologists study rocks and help to locate useful minerals. Earth scientists often work in the field—perhaps climbing mountains, exploring the seabed, crawling through caves, or wading in swamps. They measure and collect samples (such as rocks or river water), then they record their findings on charts and maps.

Explanation:

3 0
4 years ago
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