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3 years ago

What is the relationship between the mass and period in a mass hanging on spring oscillation and why?

Physics

1 answer:

Verizon [17]3 years ago

7 0

Explanation:

Period of a mass on a spring is:

T = 2π√(m/k)

T is directly proportional with the square root of m. So as the mass increases, the period increases.

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A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration

babymother [125]

acceleration = change in velocity/change in time

so...

a = 20 m/s / 2 seconds

a = 10

hope that helps :)

P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.

7 0

3 years ago

Determine the net work a hiker must do on a 3.35-kg backpack to carry it up a

Sphinxa [80]

Answer:

410.4J

Explanation:

Step one:

given

mass= 3.35kg

weight= 3.35*9.81= 32.86N

h=12.49m

Required

The net work done

Step two:

the work done is given as

WD= force* distance

WD= 32.86*12.49

WD= 410.4J

8 0

3 years ago

Hi i need answers for this. Thank you!! this is also really important and is due tomorrow at 9am!!

Allushta [10]

Answer:

You're four sentences should include about how the roller coaster has the most potential energy at the top of the track, and the opposing energy, "kinetic" has the most kinetic energy when going down the hill.

Explanation:

Kinetic - In-Motion.

Potential - Gathering Energy to go into Motion.

( I'll try to answer questions to clear up confusion. )

7 0

3 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at x = -2.7 m or at x = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

k is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the x-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be Q. It is positive.

Let the distance from charge X be x m. Then the distance from charge Y will be (0.60 - x) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at x = -2.7 m or at x = 0.27 m

3 0

3 years ago

State the role of an electromotive force (EFF) in a current flow

bonufazy [111]

Answer:

The emf source acts as a charge pump, moving negative charges from the positive terminal to the negative terminal to maintain the potential difference. This increases the potential energy of the charges and, therefore, the electric potential of the charges.

Explanation:

MARK AS BRAINLEST!!

8 0

3 years ago