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disa [49]
3 years ago
14

A horizontal spring on a frictionless surface has a spring constant of 10 N/m with a mass of 2kg attached to the end of the spri

ng. If the spring is stretched 2m passed its point of equilibrium and released, how many times does the mass pass through equilibrium per second
Physics
1 answer:
Alex3 years ago
4 0

Answer:

0.356 times the mass pass through equilibrium per second.

Explanation:

Given that,

Spring constant = 10 N/m

Mass = 2 kg

Stretched spring = 2m

We need to calculate the frequency

Using formula of frequency

f=\dfrac{1}{2pi}\sqrt{\dfrac{k}{m}}

Where, m = mass

k = spring constant

Put the value into the formula

f=\dfrac{1}{2\pi}\sqrt{\dfrac{10}{2}}

f=0.356\ Hz

We know that,

Hertz = cycle per second

Hence, 0.356 times the mass pass through equilibrium per second.

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In a calorimetry experiment, three samples A, B, and C with TA> TB> Tc are placed in thermal contact. When the samples hav
zaharov [31]

Answer:

e. TA>T>Tc

Explanation:

a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:

Q=mC_{p}(T_{f}-T_{0})

if we did an energy balance of the system, we would get that>

QA+QB+QC=0

For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.

We don't know either of them, so we cannot determine if this statement is true.

b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.

c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.

d) is not complete

e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.

3 0
3 years ago
This is the type of reaction that occurs when an acid cancels with a base.
viva [34]
Neutral reaction is the reaction that occurs when an acid cancels with a base
7 0
3 years ago
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During 9.97 seconds a motorcyclist changes his velocity from v1,x = -36.5 m/s and v1,y = 10.9 m/s to v2,x =-21.9 m/s and v2,y =
svetoff [14.1K]

Average acceleration is defined as

a = \frac{v_f - v_i}{t}

here

v_f = final\: speed

v_i = initial\: speed

t = time

now for x direction we can say

a_x = \frac{v_{2x} - v_{1x}}{t}

a_x = \frac{-21.9 + 36.5}{9.97}

a_x = 1.46 m/s^2

Similarly for Y direction

a_y = \frac{v_{2y} - v_{1y}}{t}

a_y = \frac{27.5 - 10.9}{9.97}

a_y = 1.66 m/s^2

so above is the acceleration in x and y directions

7 0
3 years ago
What role does your chosen reaction play in your life? Combustion
nirvana33 [79]

Answer:

When fuels burn in combustion reactions, they release useful thermal energy (heat). Combustion reactions are used to heat our homes, power most cars, and to generate a lot of our electricity.

Energy production from the reaction is what drives life and human greed. We stay alive by eating carbon-rich compounds, which are then burned in a spectacularly controlled and efficient manner to provide us with energy. It is the all-important energy term in the combustion equation that has made life on Earth possible

5 0
2 years ago
an 80.0 kg astronaut carrying a 15.0 kg tool kit is drifting away from the space station at a speed of 1.25 m/s. a) if she throw
jasenka [17]

The final speed of the astronaut at the space station is <u>0.36 m/s</u> and the direction will be <u>opposite</u> to the direction of the tool kit thrown.

The mass of the astronaut = 80 kg

The mass of the tool kit = 15 kg

The speed of the space station = 1.25 m/s

The speed of the tool kit threw = 6 m/s

The final speed of the astronaut can be found using the conservation of momentum formula,

           \displaystyle m_{1}v_{f1} + m_{2}v_{f2} = m_{1}v_{i1} + m_{2}v_{i2}

where m₁ is the mass of the astronaut

           m₂ is the mass of the tool kit

           \displaystyle v_{i1} and \displaystyle v_{i2} is the speed of the space station

           \displaystyle v_{f1} is the speed of the astronaut after throwing the toolkit

Let us substitute the known values in the above equation, we get

              (80 x \displaystyle v_{f1}) + (15 x 6) = (80 x 1.25) + (15 x 1.25)

                     (80 x \displaystyle v_{f1}) + 90 = 100 + 18.75

                              80 x \displaystyle v_{f1}   = 118.75 - 90

                                80 x \displaystyle v_{f1} = 28.75

                                        \displaystyle v_{f1} = 28.75 / 80

                                              = 0.36 m/s

Therefore, the final speed of the astronaut is <u>0.36 m/s</u>

By newton's third law, for every action, there will be an equal and opposite reaction. Thus, the direction of the astronaut's direction will be against the direction of the tool kit thrown.

Learn more about the conservation of momentum in

brainly.com/question/2456421

#SPJ4

5 0
1 year ago
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