Which of the following represent units of capacitance? You may choose more than one correct answer:

A particle is moving with (SHM) of period 8.0s and amplitude5.0m

nadezda [96]

Answer:

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

Explanation:

Given the description of period and amplitude, the SHM could be described by:

$f(x)=5\,sin(\frac{\pi}{4}x)$

and its angular velocity can be calculated doing the derivative:

$f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$ and is given in m/s.

Then the maximum speed is obtained when the cosine function becomes "1", and that gives:

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

The acceleration is found from the derivative of the velocity expression, and therefore given by:

$acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)$

and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

6 0

3 years ago

The construction of a flat rectangular roof (6.17 m × 5.92 m) allows it to withstand a maximum net outward force of 2.00 × 104 N

Blababa [14]

Answer:

$v_2=29.13\ m/s$

Explanation:

It is given that,

Dimension of the rectangular roof, (6.17 m × 5.92 m)

The maximum net outward force, $F=2\times 10^4\ N$

The density of air, $\rho=1.29\ kg/m^3$

The Bernoulli equation is used to find wind speed of this roof blow outward. It is given by :

$P_1+\dfrac{1}{2}\rho v_1^2=P_2+\dfrac{1}{2}\rho v_2^2$

Here, $v_1=0$ (since air inside the roof is not moving)

$v_2=\sqrt{\dfrac{2(P_1-P_2)}{\rho}}$

Since, $F=(P_1-P_2)A$

$v_2=\sqrt{\dfrac{2F}{\rho A}}$

$v_2=\sqrt{\dfrac{2\times 2\times 10^4}{1.29\times 6.17\times 5.92 }}$

$v_2=29.13\ m/s$

So, the wind speed of this roof blow outward is 29.13 m/s. Hence, this is the required solution.

5 0

4 years ago

Complete the calculations for total magnification produced by various combinations of the eyepiece and objective lenses. You may

lesya692 [45]

Answer:

a) m_ttoal = 40x , b) m_total = 100X , c) m_total = 400X ,

d) m_total = 1000 X

Explanation:

La magnificación o aumentos es el incremento de del tamaño de la imagen con respecto al tamaño original del objeto, en la mayoria del os sistema optico la magnificacion total es el producoto de la magnificación del objetivo por la magnificación del ocular

m_total = m_ objetivo * m=ocular

apliquemos esto a nuestro caso

1) m_total = 4 x * 10 x

m_ttoal = 40x

2) m_total = 10X * 10X

m_total = 100X

3)mtotal = 40X * 10X

m_total = 400X

4) m _totla = 100x * 10 X

m_total = 1000 X

en este ultimo caso para magnificación grandes es decalcificar el objeto

4 0

3 years ago

If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th

alex41 [277]

Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.
So we have 1/2 MV^2 = MGH
V^2 = 2GH
V = âš2GH
V = âš( 2 * 9.8 * 325)
V = âš 6370
V = 79.81 m/s

6 0

3 years ago

You and your friends are pushing a car with a total of 1080 N of force. You are able to push the car for a distance of 218 m. Ca

harkovskaia [24]

Answer:

How much power is required to pull a sled if you use 60J of work in 5 seconds? ... Misfortune occurs and Renatta and her friends find themselves getting a workout. They apply a cumulative force of 1080 N to push their car 218 m to the ... Calculate the amount of work done when moving a 567N crate a distance of 20 meters.

Explanation:

Misfortune occurs and Renatta and her friends find themselves getting a ... They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel ... Write down what they give you. ... Determine the work done by Lamar in deadlifting 300 kg to a height of 0.90 m ... Work = Force x Distance = Joules (force = weight).

8 0

3 years ago