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miskamm [114]
3 years ago
10

You have two springs. one has a greater spring constant than the other. you also have two objects, one with a greater mass than

the other. which object should be attached to which spring, so that the resulting spring-object system has the greatest possible period of oscillation?
Physics
1 answer:
neonofarm [45]3 years ago
4 0
The answer should be "The object with the greater mass should be attached to the spring with the smaller <span>spring constant."

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
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If the thrower takes 0.90 s to complete one revolution, starting from rest, what will be the speed of the discus at release?
Marysya12 [62]
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled

Let
α =  the angular acceleration
ω =  the final angular velocity

The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²

The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s

If the thrower's arm is r meters long, the tangential velocity of release will be 
v = 13.963r m/s

Answer: 13.963 rad/s

8 0
3 years ago
Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T
ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

                   \vec{A}=x_A\hat{i}

Vector B , which has a magnitude of 5.5 m

                   \vec{B}=x_B\hat{i}+y_B\hat{j}

                   \sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A                    \vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}

Comparing we will get

                  x_A=-x_B\\\\y_B=6x_A

Substituting in x_{B}^{2}+y_{B}^{2}=30.25

                  \left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904

So we have

    \vec{A}=0.904\hat{i}

Magnitude of vector A = 0.904

8 0
3 years ago
When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas
Bingel [31]
<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

7 0
3 years ago
A projectile is launched into the air with an initial speed of 40 m/s and a launch angle of 20° above the horizontal. The projec
miv72 [106K]

Explanation:

V=40m/s

Vy=V.sina=40.sin20=40 . 0.342=13.68m/s

Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s

Projectile travels during 5 seconds and the ramge becomes:

x=V.t=30.64 . 5=153.2m

7 0
3 years ago
Light from the sun reaches Earth in 8.3min. The speed of light is 3.00 x 10^8 m/s. How far is the Earth from the sun?
strojnjashka [21]
The answer is this, but i don't know how to simplify it. 3x^100000000<span />
5 0
3 years ago
Read 2 more answers
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