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pickupchik [31]
3 years ago
13

The motion of a pendulum for which the maximum displacement from equilibrium does not change is an example of simple harmonic mo

tion.
True or False
Physics
2 answers:
Alexeev081 [22]3 years ago
6 0
The statement "<span>The motion of a pendulum for which the maximum displacement from equilibrium does not change is an example of simple harmonic motion." is true. 

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
</span>
Alexandra [31]3 years ago
4 0

Answer:

True

Explanation:

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A mole of water molecules would just about fill a
Strike441 [17]

Answer:

B. tablespoon

Explanation:

Considering the question given, a mole of water is small compare to the options given from the question aside the table spoon.

A mole of water is just 18g and that's equivalent to 18 mL.

18 mL of water will fill a table spoon but not a cup which is about 237mL. The wheel barrow and gallon bucket have larger volumes of which 18 mL of water will never fill them.

So, a mole of water can successfully fill a table spoon.

5 0
3 years ago
Two distinct coplanar lines that do not intersect are known as
elena55 [62]

Answer:

Skew Lines

Explanation:

Two distinct lines intersect at most one point; two distinct planes intersect in at most one line. If two coplanar lines do not intersect, they are parallel/ Two lines which are not coplanar cannot intersect and are called skew lines.

7 0
3 years ago
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.1×104 m/s when at a distance
trapecia [35]

Answer:

v_f = 6.92 x 10^(4) m/s

Explanation:

From conservation of energy,

E = (1/2)mv² - GmM/r

Where M is mass of sun

Thus,

E_i = E_f will give;

(1/2)mv_i² - GmM/(r_i) = (1/2)mv_f² - GmM/(r_f)

m will cancel out to give ;

(1/2)v_i² - GM/(r_i) = (1/2)v_f² - GM/(r_f)

Let's make v_f the subject;

v_f = √[(v_i)² + 2MG((1/r_f) - (1/r_i))]

G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²

Mass of sun is 1.9891 x 10^(30) kg

v_i = 2.1×10⁴ m/s

r_i = 2.5 × 10^(11) m

r_f = 4.9 × 10^(10) m

Plugging in all these values, we have;

v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) - (1/(2.5 × 10^(11)))] 20.408 e12

v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]

v_f = √[(441000000) + (435.38 x 10^(7))

v_f = 6.92 x 10^(4) m/s

5 0
4 years ago
Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E
Leviafan [203]
We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\
Because both carts have the same mass we can cancel those out:
3v=v_A+2v_B
Now we have our system of equation that we have to solve:
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B
Part A
We need to solve our system for v_a. We will solve second equation for v_b and then plug that in the first equation.
3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}
Now we have to plug this in the first equation:
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\
We will multiply the first equation with 2 and divide by m:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\
Now we plug in the second equation into first one:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0
We end up with quadratic equation that we have to solve, I won't solve it by hand. 
Coefficients are:
a=1\\&#10;b=-6v\\&#10;c=v^2-\frac{E}{m}
Solutions are:
v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}
Part B
We do the same thing here, but we must express v_a from momentum equation:
3v=v_A+2v_B\\&#10;v_A=3v-2v_B
Now we plug this into our energy conservation equation:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\&#10;3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\&#10;3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\&#10;3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\&#10;3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\&#10;3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0&#10;
Again we end up with quadratic equation. Coefficients are:
a=3\\&#10;b=-6v\\&#10;c=6v^2-\frac{3E}{2m}
Solutions are:
v_B=\frac{6v+\sqrt{-36v^2+\frac{18E}{m}}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+\frac{18E}{m}}}{6}



8 0
3 years ago
Consider the incomplete equation below mc020-1.jpg If this equation was completed, which statement would it best support? Nuclea
vagabundo [1.1K]
<span>Nuclear fission produces elements that are heavier than helium.

The elements that are used in nuclear fission and their products are much heavier than helium.</span>
8 0
3 years ago
Read 2 more answers
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