Will give brainliest.... Physics with work please

The displacement in simple harmonic motion is a maximum when the 1. velocity is a maximum. 2. kinetic energy is a maximum. 3. ve

dlinn [17]

Answer:

3. velocity is zero.

Explanation:

The velocity of a simple harmonic motion is given by

$v = \omega\sqrt{A^2-x^2}$

Here, ω is the angular velocity, A is the amplitude (or maximum displacement from the equilibrium point) and x is the displacement at any time.

At maximum displacement, x = A. Then

$v = \omega\sqrt{A^2-A^2} = 0$

Therefore, at maximum displacement, velocity is 0.

Practically, this can be observed in a simple pendulum. As it approaches the maximum displacement, its velocity reduces. It becomes zero at this point and then reverses as the pendulum changes course. Then the velocity begins to increase. It becomes maximum at the equilibrium point but once past that, the velocity begins to reduce as it approaches the other amplitude.

For acceleration,

$a = -\omega^2x$

It follows that at maximum displacement, the acceleration is a maximum. The negative sign indicates that it is in an opposite direction to the displacement. Both kinetic energy ( $\frac{1}{2}mv^2$ ) and linear momentum ( $mv$ ) are proportional to velocity; they are therefore both zero at the maximum displacement.

5 0

3 years ago

A body of volume 100cc immersed completely in water contained in a jar. The weight of water and jar before immersion of the body

BlackZzzverrR [31]

<h2>Answer:</h2>

800gm

<h2>Explanation:</h2>

Archimedes principle states that when an object is immersed in a liquid there is an apparent loss of weight of the object. This apparent loss of weight is also the upthrust experienced by the liquid. The upthrust is equal to the weight of the liquid displaced.

Following from the above statement, when the body of volume 100c.c is immersed in the water contained in the jar, the upthrust experienced is equal to the weight of the water displaced.

Note: In the question, weight is measured just using the mass.

Mass (m) is the product of density (ρ) of liquid (which is water in this case) and volume (v) of body immersed. i.e

m = ρ x v

Where;

ρ = 1 gm/cm³

v = 100c.c = 100cm³

=> m = 1 gm/cm³ x 100cm³

=> m = 100gm

Therefore the weight of water displaced is 100gm

Now, the weight of the water and jar after immersion is the sum of the weight of water and jar before immersion, and the weight of the water displaced. i.e

Weight of water and jar after immersion = 700gm + 100gm = 800gm

8 0

3 years ago

"How did your current and voltage measurements differ between the series and parallel circuits you created

irakobra [83]

Answer:

Series circuit:

The voltage that is measured across the circuit is different.

The current measured in a series circuit remains the same at all points in the circuit.

Parallel circuit:

The current measured across each resistor varies

The voltage measured across a parallel circuit will remain the same

Explanation:

Series and parallel circuits behave differently when it comes to the circulation of current and the interaction with a potential difference.

In a series circuit, the resistances are connected end to end. As a result, the voltage that is measured across the circuit is different once resistance is encountered. However, the current measured in a series circuit remains the same at all points in the circuit.

A parallel circuit behaves in an exactly opposite manner to the series circuit. In a parallel circuit, the resistances are connected side by side. As a result of this, the current measured across each resistor varies as there are circuit branches through which electric current can flow into. On the other hand, the voltage measured across a parallel circuit will remain the same

4 0

3 years ago

Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to

Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

I will take two assumptions:

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the linear density of the stretched wire) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new charge must be 10 times the original charge, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

If you want to learn more, you can read:

brainly.com/question/14514975

6 0

3 years ago

Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o

Verdich [7]

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance = d = 96 miles

Lets call Pascal velocity V in miles per hour

Now if he increases his velocity by 50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V = d/ t₁ ⇒ 1.5 * V = 96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 ) ⇒ V * (t₁ + 16 ) = 4*d ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁ (1)

V*t₁ + 16*V = 384 (2)

We solve from equation (1) t₁ = 64/V

And by substitution in equation (2)

V * (64/V) + 16* V = 384

64 + 16 *V = 384 ⇒ 16*V = 320 ⇒ V= 320/16

V = 20 miles /sec

6 0

3 years ago