If you walk 3 meters north and then 4 meters west, what is the magnitude of your displacement?

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

How do you calculate change in position? A. initial position times two B. final position plus initial position C. final position

e-lub [12.9K]

The answer is C. Final position minus initial position.

5 0

3 years ago

While walking past a construction site, a person notices a pipe sticking out of a second floor window with water pouring out. As

tia_tia [17]

Answer:

Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.

P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0

Explanation:

P1 = P2 = atmospheric pressure so, P1 - P2 = 0

h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²

From the continuity equation for fluids

A1v1 = A2v2

v2 = A1v1/A2

Substituting into the equation above

(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho

Making A2² the subject of the formula,

A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}

The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.

Thank you for reading.

4 0

2 years ago

A steel ball of mass 50 g is rolled from the left toward a ball of lead of mass 500 g. The steel ball is traveling with a veloci

alina1380 [7]

Answer: It's hard to say without characterizing the collision. But it will be either A if the collision is totally in-elastic, or B if the collision is totally elastic. It could be anywhere in between for partially elastic collisions.

Explanation:

momentum is conserved, so initial system momentum will be left to right.

The velocity of the center of mass is 50(5) / 550 = 0.4545... m/s

In an elastic collision, the lead ball will move off at twice that speed or 0.91 m/s to the right.

The steel ball will bounce back and move away at 0.91 - 5 = -4.1 m/s . The negative sign indicates the steel ball has reversed course and has negative momentum

In a totally in-elastic collision, both balls would move to the right at 0.45 m/s. The steel ball will still have positive momentum.

4 0

2 years ago

We know today that atoms cannot be divided into smaller parts true or false

djverab [1.8K]

Hi , the answer is false ,atoms can be divided into smaller parts , electrons , protons and neutrons.

3 0

2 years ago