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3 years ago

Part D

Physics

1 answer:

seropon [69]3 years ago

3 0

Answer:

Which item fits in the left side of the table?Criminal versus Civil Cases Criminal cases Civil cases ? . • Accuser is called the plaintiff • Lower burden of proof Defendant may be found liable • Usually heard by a judge only • Accused not guaranteed an attorney

A. No burden of proof

B. Accuser is called the defendant

C. Defendant may be found guilty

D. Usually heard by a jury only

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How is the egg transferred from the ovary to the uterus? through muscle contractions by using special hairs by producing estroge

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The egg is transferred from the ovary to the uterus through muscle contractions called uterine contractions by using special hair like structures called cilia.

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A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago

Give 5 examples of physical properties that could be used to describe a solid

ycow [4]

1.) appearance
2.)texture
3.)color
4.)melting point
5.)odor

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Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

Tamiku [17]

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am 1 m away from B , the path difference will be

8 - 1 = 7 m which is odd multiple of 1 or λ /2

So path difference becomes odd multiple of λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

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How many significant digits are in the number 0.010?

Soloha48 [4]

There a two significant digits

explanation:
Trailing zeros after a decimal point count if preceded by a non-zero value. Example: 0.01 one significant figure, 0.010 two significant figures, 0.0100 three significant figures.

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