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3 years ago

13

One cell performs 120 oscillations per minute, the other 240. What will be the ratio of their lengths ( I translated it sorry fo

r mistakes)

1 answer:

faltersainse [42]3 years ago

6 0

Answer:

360²

Explanation:

hoped this helped (ejfigueroa5 had the answer thank you to ejfigueroa5)

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Which branch of physics deals with the study of force, energy, and motion?

yawa3891 [41]

The branch of physics that deals with the study of force energy and motion is classic mechanics

8 0

3 years ago

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 $1$ .

So

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

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 $=0.33\ m$ .

3 0

3 years ago

A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring

AleksandrR [38]

Answer:

The magnitude of the horizontal net force is 13244 N.

Explanation:

Given that,

Mass of car = 1400 kg

Speed = 17.7 m/s

Distance = 33.1 m

We need to calculate the acceleration

Using equation of motion

$v^2-u^2=2as$

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

$0-(17.7)^2=2a\times 33.1$

$a=\dfrac{-(17.7)^2}{33.1}$

$a=-9.46\ m/s^2$

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

$F = ma$

$F =1400\times(-9.46)$

$F=-13244\ N$

Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

$|F| =13244\ N$

Hence, The magnitude of the horizontal net force is 13244 N.

5 0

4 years ago

Plz help I will mark brainiest help send answers

Elden [556K]

Answer:

D

Explanation:

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3 years ago

6) what is ararge of speed

irina [24]

Answer:

in everyday use and in kinematic the speed of an object is the magnitude of the change of its position it is thus a scalar quantity

5 0

3 years ago