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3 years ago

Which quantity measures the maximum distance an object on a spring moves from the equilibrium position during oscillation?

Physics

2 answers:

nirvana33 [79]3 years ago

7 0

Answer:

amplitude

Explanation:

The maximum distance from the mean position or the equilibrium position on its either side of an oscillating body is called its amplitude.

nasty-shy [4]3 years ago

5 0

The appropriate response is amplitude. A measure of its change over a solitary period. There are different meanings of plentifulness, which are all elements of the extent of the distinction between the variable's outrageous qualities. In more seasoned writings, the stage is now and again called the adequacy.

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When a wire loop is connected to a battery, what is produced in the loop?

Vinil7 [7]

Answer:

magnetic field

Explanation:

3 0

3 years ago

What does vf stand for<br> a.fringe velocity<br> b.first velocity<br> c.final velocity

Elza [17]

The correct answer is C. Final Velocity

Hope this helped!

5 0

3 years ago

Explain how the basic units are combined to give the derived units of force, velocity and pressure and work

Lorico [155]

#Force

$\\ \sf\longmapsto Force=Mass\times Acceleration$

$\\ \sf\longmapsto Force=kg m/s^2$

$\\ \sf\longmapsto Force=Newton (N)$

#Velocity

$\\ \sf\longmapsto Velocity=\dfrac{Displacement}{time}$

$\\ \sf\longmapsto Velocity=m/s$

#Work

$\\ \sf\longmapsto Work=Force\times Displacement$

$\\ \sf\longmapsto Work=Nm$

$\\ \sf\longmapsto Work=Joules(J)$

5 0

2 years ago

An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 25.0° above

Greeley [361]

Answer:

$v_{i}= 19\times 10^5\ m/s$

Explanation:

given,

scattering angle of alpha particle = 25.0° above its initial direction of motion

oxygen nucleus recoils at = 50.0° below this initial direction.

final speed of the oxygen = 2.08×10⁵ m/s

mass of alpha particle = 4.0 u

mass o oxygen nucleus = 16 u

momentum conservation along x- axis

$m_{a}v_{i} = m_a v_a cos\theta + m_o v_o cos\theta$

$4v_{i} = 4\times v_a cos25^0 + 16\times 2.08 \times 10^5 cos50^0$

$v_{i}= \dfrac{3.625\times v_a+ 21.39\times 10^5}{4}$ ....(1)

Along y-direction

$0 = m_av_a sin \theta - m_ov_o sin\theta$

$0 = 4\times v_a sin 25 - 16\times 2.08 \times 10^5 sin50^0$

$v_a = \dfrac{25.49 \times 10^5}{1.69}$

$v_a = 15.082\times 10^5\ m/s$

putting value in equation (1)

$v_{i}= \dfrac{3.625\times 15.082\times 10^5+ 21.39\times 10^5}{4}$

$v_{i}= 19\times 10^5\ m/s$

5 0

3 years ago

A 2.7-kg cart is rolling along a frictionless, horizontal track towards a 1.1-kg cart that is held initially at rest. The carts

olga2289 [7]

Answer:

Part a)

P = 8.23 kg m/s

Part b)

v = 3.05 m/s

Explanation:

Part a)

momentum of cart 1 is given as

$P_1 = m_1v_1$

$P_1 = (2.7)(3.7) = 9.99 kg m/s$

Momentum of cart 2 is given as

$P_2 = m_2v_2$

$P_2 = (1.1)(-1.6) = -1.76 kg m/s$

Now total momentum of both carts is given as

$P = P_1 + P_2$

$P = 8.23 kg m/s$

Part b)

Since two carts are moving towards each other due to mutual attraction force and there is no external force on two carts so here momentum is always conserved

so here we will have

$P_i = P_f$

$(2.7 kg)v = 8.23$

$v = 3.05 m/s$

3 0

3 years ago